Indefinite integral

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Definite and indefinite integrals

It is time to get familiar with computing integrals. There are two types: definite and indefinite integrals.

Definite integration deals with calculating the area under the curve of a function.

Here is a function

and the area under its curve from a to b is:

Indefinite integration works a totally different way.

The reason we call it indefinite is that it does not have a and b limits for the integration, we just simply integrate the function:

The indefinite integral of f(x) is a function that is called the primitive function.

The symbol for the primitive function is F(x), and its main property is that if we differentiate it, we get f(x).

So, this indefinite integration is nothing else than the reverse of differentiation.

This is why it is sometimes called antidifferentiation.

Let's see a few examples.

Here is this one, for instance:

We need a function whose derivative is 2x.

There is such a function, namely

Here comes another one:

There is also a function, whose derivative is

If we can recall,

If one knows what absolute value is, it won't be a very disturbing piece of news that we need to use that here, too. The reason is that we want to integrate

for negative x values, too.

But lnx only likes positive x values. Luckily, the absolute value solves this minor problem.

However, it is enough to remember this:

Finally, let's see one more:

What should we differentiate to get x2?

This is almost perfect; we just have to divide it by 3.

And just one more thing. If we differentiate x2, of course it will be 2x, but

This means any constant can be added to x2.

And here too.

And now, let's see the relationship between definite and indefinite integrals.

The theorem that describes this relationship is one of the most important theorems in the entire history of mathematics.

It was developed at the end of the 1600s by two men at the same time: an English physicist named Newton and a German philosopher named Leibniz.

If f(x) is integrable on the closed interval [a, b], and it has a primitive function on this interval, then

This thing here means the change of the primitive function, so first we substitute b, then we substitute a, and then subtract it.

Let's try how this theorem works: let's find the area under the curve of x2 from 0 to 1.

Here comes the primitive function, and we have to find its change from 0 to 1.

If we can't remember the primitive function, then we are in trouble.

Let's compute the area under the curve of

from 0 to 1.

It isn’t particularly challenging to write the formula of what we should integrate.

However, we have no idea what the primitive function is, and that is a showstopper.

So, the crux of the matter is finding the primitive function, in other words, finding the indefinite integral.

We can’t push it off anymore: it is time to master that skill.


Basic integration formulas

We start looking for primitive functions by remembering the derivatives of a few important functions.

The first one is xn

Differentiation decreases the exponent by 1. Integration increases it by 1.

Although it's a little troublesome, if

But here comes the solution.

Then, finally an anchor point in our lives.

The list is going to be fairly long.

And this is only the beginning.

And now, we have to clarify a few very important things.

Here is one of them:

 but  

And here is the other one:

Let's try to figure out this one:

It seems logical that

Only there is a little trouble.

Integration is the reverse of differentiation, so if we integrate a function and then differentiate it, we should get back exactly to the original function.

But that doesn't seem to be true here.

We don't arrive back to the original function, because differentiation inserts this multiplier of 3 here.

But this can be helped.

If the exponent is some sort of ax+b type polynomial,

then when we integrate, we have to multiply by  .

Let's see this one, for instance:

Here, instead of being in the exponent, the ax+b polynomial is in the denominator.

We will need this method quite often, so we should store it in a handy place in our head.

Now comes the real thrill!


Integration rules

Integration is a much more entertaining activity than differentiation.

For example, here is a product:

Differentiation is very simple: we only have to remember one rule, and then we can use it for any kind of product.

But if we have to integrate, well, the situation is a lot more exciting.

There will be at least five different rules, and we will have to be able to figure out which method is needed for a given case.

A tiny change is enough to require a different method.

So, we won't get bored.

In order to overcome these minor difficulties, we group the integration rules so that we start from the simplest rule, and progress towards the more difficult ones.

Let's see the rules.

INTEGRATION RULES

The first rule is exactly like the one for differentiation. A constant multiplier can be factored out.

The second rule is about integrating sums. This one is also similar to differentiation: parts of a sum are integrated separately.

The rule for products is more exciting. Because there is no such rule.

If we have to integrate a product, we have several choices. Exactly five.

But we will have to know which rule to use, and it is not recommended to decide by tossing a coin.

Luckily, there will be a separate slideshow for each method that will reveal how to use that particular rule and when to use it.

Fractions will be a similar story. There will be a few methods for integrating fractions. Let's see, maybe three?

And composite functions also have at least two methods.

So we better get to work.

Integration rules, Constant multiplier, Integrating sums, Integrating products, Integrating a fractions, Integrating composite functions.


Integrating simple products - S1

If the multiplication can be completed, then we should do that, and then do the integration

Well, this wasn't the most difficult integration of our lives.

And this one won't be either:

It's time to see something more interesting.

We are ready for the next integration rule.


Integrating uglier products - S2

Let's see an example of this.

Here comes another one.

And a third one.

Sometimes we have to invest some energy to get everything right.

Here is this one, for instance:

Let's start identifying:

It seems that we need a multiplier of 6 here.

So, we will multiply by 6 within the integral, and divide by 6 in front of it.

Let's see what else could possibly come.

We have to improve this a bit, and then voila! Finished.

And then we have this.

There are some quite tricky cases, too.

When fα is in the denominator, first we have to ask it to please come up from there,

and then we can use the formula.

Here is another one.

And here comes one more.

Finally, let's see some magic tricks.


Integration by parts 1.0 - S3

Partial integration was developed for integrating products.

The name comes from the fact that we will integrate the product in parts.

Let's see how.

We need to do some casting.

One factor of the product will be f, and the other one will be g'.

But it is not recommended to decide by tossing a coin.

So, it would be useful to have an idea how to pick the roles.

Let's try it this way first:

The purpose of partial integration is to turn a complicated integral into a simpler one.

The casting is good, when we achieve this goal.

This time we failed.

This integral is even more complicated than the original, because instead of x, it contains x2 .

What did we mess up?

Well, the thing is, there are two open roles, f and g'.

During the procedure, we differentiate f, integrate g', and these will appear in the new integral.

If we assign the role of g' to x, then when we integrate, we will increase x's exponent.

If we assign the role of f to x, then we will differentiate it, and x's exponent will be decreased.

We want to decrease the exponent.

Thus, we have to reverse the roles.

With this casting, we differentiate x, and it becomes 1.

For ex, it doesn't matter, it becomes ex, just like before.

Here, we achieved our goal this second integral is indeed simpler.

Let's see one more:

Based on the previous discussions, it seems that f=x2 will be the winning choice.

We achieved our goal: successfully turned a complicated integral into a simpler one.

But it is still not simple enough, so we do another partial integration.

We learned so far that it seems best to assign the role of f to the power of x in the product.

Let's make a note of this for ourselves.

CASTING:

But, life would be too easy without some annoying exceptions.

Here is this one, for instance:

Based on our notes, we should assign the roles like this:

But unfortunately, it leads to nothing. If you don't believe it, you can try and see yourself.

So, we reverse the roles.

We extend our list.

Cases where we should use reverse casting:

And there is one more thing.

Here are these functions from cases where reverse casting is necessary.

Well, let's try to integrate them.

We will need a bit of a trick.

We will use partial integration. Based on our list, we will call these f, and here comes the trick: g’=1

Let's see another one, too.

The purpose of partial integration is to turn a complicated integral into a simpler one.

In the previous slideshow we saw how partial integration works, and now we will solve a few splendid problems.

Here comes this one, for instance:

¶=f  · g- f f’·g

Let's see one more:

                                             f f·g’ =f·g- f f’·g

For partial integration, we assign the roles:

 =f  · g- f f’·g

Here comes another one.

Well, there is such a thing as:

We should be careful with this partial integration thing, as it could be harmful in large doses.

But perhaps one more is safe.

Well, there is this thing, too:

And this one, too:

  =f  · g- f f’·g

The point of partial integration is that it turns complicated integrals into simpler ones.

The process that transforms a more complicated integral into a simpler integral is called reduction.

Here come a few reduction formulae. There is really no point in memorizing these, as we can calculate them at any time.

We can use these to finish off a few of the nastier integrals.


Integration by parts 2.0 - S3

Partial integration was developed for integrating products.

The name comes from the fact that we will integrate the product in parts.

Let's see how.

We need to do some casting.

One factor of the product will be f, and the other one will be g'.

But it is not recommended to decide by tossing a coin.

So, it would be useful to have an idea how to pick the roles.

Let's try it this way first:

The purpose of partial integration is to turn a complicated integral into a simpler one.

The casting is good, when we achieve this goal.

This time we failed.

This integral is even more complicated than the original, because instead of x, it contains x2 .

What did we mess up?

Well, the thing is, there are two open roles, f and g'.

During the procedure, we differentiate f, integrate g', and these will appear in the new integral.

If we assign the role of g' to x, then when we integrate, we will increase x's exponent.

If we assign the role of f to x, then we will differentiate it, and x's exponent will be decreased.

We want to decrease the exponent.

Thus, we have to reverse the roles.

With this casting, we differentiate x, and it becomes 1.

For ex, it doesn't matter, it becomes ex, just like before.

Here, we achieved our goal this second integral is indeed simpler.

Let's see one more:

Based on the previous discussions, it seems that f=x2 will be the winning choice.

We achieved our goal: successfully turned a complicated integral into a simpler one.

But it is still not simple enough, so we do another partial integration.

We learned so far that it seems best to assign the role of f to the power of x in the product.

Let's make a note of this for ourselves.

CASTING:

But, life would be too easy without some annoying exceptions.

Here is this one, for instance:

Based on our notes, we should assign the roles like this:

But unfortunately, it leads to nothing. If you don't believe it, you can try and see yourself.

So, we reverse the roles.

We extend our list.

Cases where we should use reverse casting:

And there is one more thing.

Here are these functions from cases where reverse casting is necessary.

Well, let's try to integrate them.

We will need a bit of a trick.

We will use partial integration. Based on our list, we will call these f, and here comes the trick: g’=1

Let's see another one, too.

The purpose of partial integration is to turn a complicated integral into a simpler one.

In the previous slideshow we saw how partial integration works, and now we will solve a few splendid problems.

Here comes this one, for instance:

¶=f  · g- f f’·g

Let's see one more:

                                             f f·g’ =f·g- f f’·g

For partial integration, we assign the roles:

 =f  · g- f f’·g

Here comes another one.

Well, there is such a thing as:

We should be careful with this partial integration thing, as it could be harmful in large doses.

But perhaps one more is safe.

Well, there is this thing, too:

And this one, too:

  =f  · g- f f’·g

The point of partial integration is that it turns complicated integrals into simpler ones.

The process that transforms a more complicated integral into a simpler integral is called reduction.

Here come a few reduction formulae. There is really no point in memorizing these, as we can calculate them at any time.

We can use these to finish off a few of the nastier integrals.


Integrating composite functions - S4

This theorem is really about integrating composite functions. But unfortunately composite functions are a bit problematic, as computing their integral is usually a quite hopeless undertaking.

There is no elementary primitive function for any of these functions:

Therefore sadly, we cannot compute these integrals. I don't mean today, but at all. 

We may have some hope, if these functions are multiplied by the derivative of their internal functions.

It is worth to remember a few special cases:

Here are a few exercises for them.

There are cases where we have to invest some energy to get everything right.

There are usually two possibilities.

The easier case is when the function we need to integrate differs from the ideal situation only by a constant. The other case is when the factors containing x are also different.

If the difference is only in a constant, then it can be resolved easily:

EXAMPLES:

The other case is much more unpleasant. Let's see an example for that!

At first sight, it looks like an

type case, but there is a little trouble.

Here, we need to have the derivative of the exponent, but the derivative of x2 is 2x.

Here comes an idea: if we need 2x there, then well, let's write 2x there.

But that would change the problem. To prevent the multiplication from changing the problem, we should also divide by 2x.

So, we multiplied by 2x, but also divided by 2x, thus the original problem did not change.

On the other hand, here appeared the derivative of the exponent, so now we can integrate.

The only question is what to do with this part.

We will integration by parts.

A bit more integration here, and then it's done.


Here they come the fractions - T1

Let's try the partial fraction decomposition, and then the integration.

The partial fraction method is useful when the denominator consists of only one member, in other words, if there is no addition in the denominator.

But sometimes it may work, even if there is an addition.

And then there are some quite tricky cases, too.

And, that's all about partial fraction decomposition.


An important rule for fractions - T2

T2

Let's see an example of this.

differentiate

Here comes another one:

Sometimes the numerator is not exactly the derivative of the denominator, but very similar.

In such cases, to draw inspiration, we differentiate the denominator, and compare it with the numerator to see what we should do to succeed.

There is, sadly only x in the numerator, while we would need 4x.

So, we multiply the numerator by 4, and in turn, we divide by 4 outside of the integral sign.

Having fewer x in the numerator can cause trouble, and so does having more.

.

The derivative of the denominator is , but in the numerator we have , which is more than we need, so we factor out a 3.

Finally, there are some very cunning cases, too.

There are some that are even trickier.


Integration by substitution 1.0

INTEGRATION BY SUBSTITUTION

The main idea of integration by substitution is that we replace an expression by u, hoping that this way we may be able to solve the problem.

Let's see an example for this!

In such cases it is best to assign u to be the entire square root expression.

USEFUL SUBSTITUTIONS

So far so good, now we solve for x.

And presumably, we would later substitute that here.

But sadly, there is one more thing here.

Well, dx should be replaced, too, the following way:

We differentiate the expressed x with respect to u.

If what you see here is surprising to you, well, there is an explanation for everything.

Long-long time ago, people did not use the well-known f’ notation for differentiation. Instead, they used

Some people still use this notation.

Later the notations were simplified, but for some mysterious reason, here, at

integration by substitution, this ancient method was kept.

So, let's resign ourselves to this, the derivative of the expressed x is not the usual

but

Let's hope that nobody would lose sleep over this.

And now comes the substitution.

Well, this seems to be done.

Let's see another one like this.

Integration by substitution is also useful for dealing with ugly composite functions.

In such cases we usually assign u to the internal function.

Now, we integrate by parts.

Let's see another one like this.

We have to express x from this

using a little trick.

There is such a thing as:

so this one is done.

We only have to differentiate with respect to u.

And here comes the substitution.

Now, we integrate. The derivative of the denominator

happens to be the numerator.

There are more exciting substitutions, too.

They will appear in the next slideshow. But only for those who like such thrills.

Integration by substitution works so that we replace an expression by u, hoping that this way we may be able to solve the problem.

So far, we worked on easier substitutions, now we are ready for more difficult ones. Let's start by looking at some square root cases.

When the expression under the square root sign is linear, that is the more pleasant case, and it is best to substitute u for the entire square root expression.

If, however, the expression under the square root sign is non-linear, then we are dealing with an entirely different animal.

In such cases it doesn't always worth to try substitution.

It is quite suspicious that the derivative of the1 – x4 under the square root sign is almost the same as the numerator.

So, we do not substitute, but instead

use a method called S2.

But if, annoyingly instead of x3, there is x2 in the numerator,

then the previous method doesn't work.

So, in our despair, we opt for integration by substitution.

Here is the menu.

This is exactly what we need now.

Let's see where this takes us.

Now come where we differentiate x.

First, we differentiate the external function,

and then the internal.

So far, this does not look very encouraging. But let's see the substitution.

And now comes the main point.

We do this whole substitution business, because

and this way we can get rid of the square root sign.

There is also the added benefit of being able to simplify.

At the end we realize that this is the easiest integral of our lives.

All we have left to do is replacing u with whatever we called u at the beginning.

Wait, what did we call u?

Now we take the arcsinus of both sides.

This is the inverse of sinus, thus

Let's see another exciting case.

Here is where we get rid of the square root sign.

And now it would be good to integrate.

We need a bit of a trick for that.

This one again…

 Then we do some cutting (T1)

Finally, we replace whatever we called u.

The next story is about substituting trigonometric functions.

Let's see a square root case in this j(t) internal function version.

In such cases it is best to assign t to the entire square root expression in the denominator.

So,  , thus  and .

This latter one will be the j(t) function,

in other words,  and .

After completing the substitution, the problem looks like this:

Here, we have to substitute for t, and then it is done.


Integration by substitution 2.0

INTEGRATION BY SUBSTITUTION

The main idea of integration by substitution is that we replace an expression by u, hoping that this way we may be able to solve the problem.

Let's see an example for this!

In such cases it is best to assign u to be the entire square root expression.

USEFUL SUBSTITUTIONS

So far so good, now we solve for x.

And presumably, we would later substitute that here.

But sadly, there is one more thing here.

Well, dx should be replaced, too, the following way:

We differentiate the expressed x with respect to u.

If what you see here is surprising to you, well, there is an explanation for everything.

Long-long time ago, people did not use the well-known f’ notation for differentiation. Instead, they used

Some people still use this notation.

Later the notations were simplified, but for some mysterious reason, here, at

integration by substitution, this ancient method was kept.

So, let's resign ourselves to this, the derivative of the expressed x is not the usual

but

Let's hope that nobody would lose sleep over this.

And now comes the substitution.

Well, this seems to be done.

Let's see another one like this.

Integration by substitution is also useful for dealing with ugly composite functions.

In such cases we usually assign u to the internal function.

Now, we integrate by parts.

Let's see another one like this.

We have to express x from this

using a little trick.

There is such a thing as:

so this one is done.

We only have to differentiate with respect to u.

And here comes the substitution.

Now, we integrate. The derivative of the denominator

happens to be the numerator.

There are more exciting substitutions, too.

They will appear in the next slideshow. But only for those who like such thrills.

Integration by substitution works so that we replace an expression by u, hoping that this way we may be able to solve the problem.

So far, we worked on easier substitutions, now we are ready for more difficult ones. Let's start by looking at some square root cases.

When the expression under the square root sign is linear, that is the more pleasant case, and it is best to substitute u for the entire square root expression.

If, however, the expression under the square root sign is non-linear, then we are dealing with an entirely different animal.

In such cases it doesn't always worth to try substitution.

It is quite suspicious that the derivative of the1 – x4 under the square root sign is almost the same as the numerator.

So, we do not substitute, but instead

use a method called S2.

But if, annoyingly instead of x3, there is x2 in the numerator,

then the previous method doesn't work.

So, in our despair, we opt for integration by substitution.

Here is the menu.

This is exactly what we need now.

Let's see where this takes us.

Now come where we differentiate x.

First, we differentiate the external function,

and then the internal.

So far, this does not look very encouraging. But let's see the substitution.

And now comes the main point.

We do this whole substitution business, because

and this way we can get rid of the square root sign.

There is also the added benefit of being able to simplify.

At the end we realize that this is the easiest integral of our lives.

All we have left to do is replacing u with whatever we called u at the beginning.

Wait, what did we call u?

Now we take the arcsinus of both sides.

This is the inverse of sinus, thus

Let's see another exciting case.

Here is where we get rid of the square root sign.

And now it would be good to integrate.

We need a bit of a trick for that.

This one again…

 Then we do some cutting (T1)

Finally, we replace whatever we called u.

The next story is about substituting trigonometric functions.

Let's see a square root case in this j(t) internal function version.

In such cases it is best to assign t to the entire square root expression in the denominator.

So,  , thus  and .

This latter one will be the j(t) function,

in other words,  and .

After completing the substitution, the problem looks like this:

Here, we have to substitute for t, and then it is done.


Integration by substitution 3.0

INTEGRATION INVOLVING COMBINATIONS OF TRIGONOMETRIC FUNCTIONS

Integration of trigonometric terms is not easy. We will check out a few simpler tricks and the most important methods. Let's start with one of the most exciting methods.

One of the strangest cases of integration by substitution is when u= tan(x/2).

We use this if sinx and cosx is included in first-degree form in the fraction.

The main idea of the substitution is the following three formulae:

Now we will see some magic tricks.

Those who are not particularly fond of magic, can skip this part.

We start with a smaller trick:

Then, we use a more advanced trick.

At the end comes a finishing trick.

The hocus-pocus ends here, and we got this:

And, one more thing.

For the substitution, we need the derivative of the expressed x, too.

So, we need to express x first.

There is such a thing as:

This way, bye-bye tangent, we found x.

And we found the derivative, too. We know that

The good news is that we had to suffer through this previous ordeal only this one time.

From here, all we have to do is make a note of these, so we could refer to them when needed.

Here comes an exercise.

This method works in more complicated cases, too:

We do some decomposition

 and then it's done.

And that's all I have to say about that.


Integrating rational functions 1.0

INTEGRATING RATIONAL FUNCTIONS

Integrating rational functions is an extremely entertaining activity.

We will start by developing our ability to integrate so called partial fractions.

There are two kinds of partial fractions:

I.                                                                           II.

In the first type of partial fraction the denominator is a linear polynomial, and the numerator is a constant.

In the second type of partial fraction the denominator is an irreducible quadratic polynomial, and the numerator is a linear polynomial.

Let's see how we integrate partial fractions.

Then, there is this thing:

We modify the numerator a bit, such that the derivative of the denominator appears in it.

This is needed here, so we add it and also subtract it.

And here, the derivative of the denominator appeared in the numerator, accompanied by some constant.

Now we decompose the fraction into the sum of two fractions:

The first integral is done:

We have to suffer a bit more with the other one.

We create a perfect square in the denominator.

Here, in the denominator, a perfect square appears. For the sake of simplicity, we will call the term behind it D.

We call these D, and voila!:

Now we should solve a problem.

We can easily integrate any rational function. All we have to do is decompose it into partial fractions, and then integrate the partial fractions using the previous methods.

There happens to be a problem here:

First let's check whether the degree of the numerator is less than that of the denominator. If this does not hold, then we will need to use polynomial long division.

The polynomial division is wonderful, we will check it out later, but luckily we don't need it now.

It is because the numerator is quadratic, and the denominator is cubic.

Let's start to decompose into partial fractions. For that, we need to factorize the denominator into linear and irreducible quadratic terms.

We can factor out x,

and this cannot be factorized further,

as it has a negative discriminant.

With this, factorization is done.

These are going to be the denominators of the partial fractions.

Now we have to figure out the numerators. Not the actual numerators just yet, but their parametric forms. Let's see what this means.

We are determined to decompose into partial fractions. And there are two types of partial fractions.

Type I partial fractions have a linear denominator.

Type II partial fractions have a non-factorable quadratic denominator.

During decomposition we always start with the denominators. The denominator of the first fraction is clearly linear, so it could only be a Type I partial fraction. So the numerator is some A.

On the other hand, the denominator of the second fraction is a quadratic polynomial, so this fraction is Type II, and thus its numerator is in the form of Ax+B.

Well, A is already in use, so let's say Bx+C will be the numerator.

And now we calculate what A, B and C are.

For this, we have to stare at them a bit.

We multiply by the denominators,

and then expand the parentheses.

And then we check how many x2 terms, x terms and constants are on the right side.

Because there are exactly the same on the left side as well.

We solve the system of equations.

The first fraction is done, the second will take a bit more time.

First we create the derivative of the denominator, then

We can easily integrate any rational function. All we have to do is decompose it into partial fractions, and then integrate the partial fractions using the previous methods.

There happens to be a problem here:

First let's check whether the degree of the numerator is less than that of the denominator. If this does not hold, then we will need to use polynomial long division.

The polynomial division is wonderful, we will check it out later, but luckily we don't need it now.

Let's factorize the denominator. Next, factor out x.

Then let's see whether the quadratic factor could be further factorized.

It seems, yes. If someone doesn't feel like doing such factorization, well, here is a splendid little formula:

where

Factorization is done. These are going to be the denominators of the partial fractions.

We figure out the numerators based on the denominators.

Since all three denominators are linear, all three factors are of Type I partial fractions, so the numerators are A, B and C.

Now we calculate what A, B and C are.

For this, we have to stare at them a bit.

We multiply by the denominators,

and then comes a trick.

Let's see what happens if we substitute zero for x.

Now let's try to compute the value of B.

For that, we should make these zero.

Finally, to compute C, we will make these zero.

If you did not like the trick, we could also expand the parentheses:

And then we check how many x2 terms, x terms and constants are on the right side.

Because there are exactly the same on the left side as well.

We solve the system of equations.

Here is another rational function:

Again, we need to factorize the denominator into linear and irreducible quadratic terms.

Let's see if this could be factorized.

Well, it seems, yes.

Now come the decomposing into partial fractions. As we can see, the denominator contains one of the linear factors twice. In such cases there is a small trick for the decomposition.

The denominator of one partial fraction is (2x+1), and the other's is (2x+1)2.

Again, we figure out the numerators based on the denominators.

Since all three denominators are linear, or a power of a linear term, all three factors are of Type I partial fractions, so the numerators are A, B and C.

Now let's see what A, B and C are.

We will use the tricky method shown in the previous slideshow.

First we zero these:

These cannot be zero at the same time, so finding A will be a bit more difficult.

Well, let's substitute 0 for x.

We could substitute 666, too but 0 is safer.

These are easy to integrate.


Integrating rational functions 2.0

INTEGRATING RATIONAL FUNCTIONS

Integrating rational functions is an extremely entertaining activity.

We will start by developing our ability to integrate so called partial fractions.

There are two kinds of partial fractions:

I.                                                                           II.

In the first type of partial fraction the denominator is a linear polynomial, and the numerator is a constant.

In the second type of partial fraction the denominator is an irreducible quadratic polynomial, and the numerator is a linear polynomial.

Let's see how we integrate partial fractions.

Then, there is this thing:

We modify the numerator a bit, such that the derivative of the denominator appears in it.

This is needed here, so we add it and also subtract it.

And here, the derivative of the denominator appeared in the numerator, accompanied by some constant.

Now we decompose the fraction into the sum of two fractions:

The first integral is done:

We have to suffer a bit more with the other one.

We create a perfect square in the denominator.

Here, in the denominator, a perfect square appears. For the sake of simplicity, we will call the term behind it D.

We call these D, and voila!:

Now we should solve a problem.

We can easily integrate any rational function. All we have to do is decompose it into partial fractions, and then integrate the partial fractions using the previous methods.

There happens to be a problem here:

First let's check whether the degree of the numerator is less than that of the denominator. If this does not hold, then we will need to use polynomial long division.

The polynomial division is wonderful, we will check it out later, but luckily we don't need it now.

It is because the numerator is quadratic, and the denominator is cubic.

Let's start to decompose into partial fractions. For that, we need to factorize the denominator into linear and irreducible quadratic terms.

We can factor out x,

and this cannot be factorized further,

as it has a negative discriminant.

With this, factorization is done.

These are going to be the denominators of the partial fractions.

Now we have to figure out the numerators. Not the actual numerators just yet, but their parametric forms. Let's see what this means.

We are determined to decompose into partial fractions. And there are two types of partial fractions.

Type I partial fractions have a linear denominator.

Type II partial fractions have a non-factorable quadratic denominator.

During decomposition we always start with the denominators. The denominator of the first fraction is clearly linear, so it could only be a Type I partial fraction. So the numerator is some A.

On the other hand, the denominator of the second fraction is a quadratic polynomial, so this fraction is Type II, and thus its numerator is in the form of Ax+B.

Well, A is already in use, so let's say Bx+C will be the numerator.

And now we calculate what A, B and C are.

For this, we have to stare at them a bit.

We multiply by the denominators,

and then expand the parentheses.

And then we check how many x2 terms, x terms and constants are on the right side.

Because there are exactly the same on the left side as well.

We solve the system of equations.

The first fraction is done, the second will take a bit more time.

First we create the derivative of the denominator, then

We can easily integrate any rational function. All we have to do is decompose it into partial fractions, and then integrate the partial fractions using the previous methods.

There happens to be a problem here:

First let's check whether the degree of the numerator is less than that of the denominator. If this does not hold, then we will need to use polynomial long division.

The polynomial division is wonderful, we will check it out later, but luckily we don't need it now.

Let's factorize the denominator. Next, factor out x.

Then let's see whether the quadratic factor could be further factorized.

It seems, yes. If someone doesn't feel like doing such factorization, well, here is a splendid little formula:

where

Factorization is done. These are going to be the denominators of the partial fractions.

We figure out the numerators based on the denominators.

Since all three denominators are linear, all three factors are of Type I partial fractions, so the numerators are A, B and C.

Now we calculate what A, B and C are.

For this, we have to stare at them a bit.

We multiply by the denominators,

and then comes a trick.

Let's see what happens if we substitute zero for x.

Now let's try to compute the value of B.

For that, we should make these zero.

Finally, to compute C, we will make these zero.

If you did not like the trick, we could also expand the parentheses:

And then we check how many x2 terms, x terms and constants are on the right side.

Because there are exactly the same on the left side as well.

We solve the system of equations.

Here is another rational function:

Again, we need to factorize the denominator into linear and irreducible quadratic terms.

Let's see if this could be factorized.

Well, it seems, yes.

Now come the decomposing into partial fractions. As we can see, the denominator contains one of the linear factors twice. In such cases there is a small trick for the decomposition.

The denominator of one partial fraction is (2x+1), and the other's is (2x+1)2.

Again, we figure out the numerators based on the denominators.

Since all three denominators are linear, or a power of a linear term, all three factors are of Type I partial fractions, so the numerators are A, B and C.

Now let's see what A, B and C are.

We will use the tricky method shown in the previous slideshow.

First we zero these:

These cannot be zero at the same time, so finding A will be a bit more difficult.

Well, let's substitute 0 for x.

We could substitute 666, too but 0 is safer.

These are easy to integrate.


Integrating rational functions 3.0

INTEGRATING RATIONAL FUNCTIONS

Integrating rational functions is an extremely entertaining activity.

We will start by developing our ability to integrate so called partial fractions.

There are two kinds of partial fractions:

I.                                                                           II.

In the first type of partial fraction the denominator is a linear polynomial, and the numerator is a constant.

In the second type of partial fraction the denominator is an irreducible quadratic polynomial, and the numerator is a linear polynomial.

Let's see how we integrate partial fractions.

Then, there is this thing:

We modify the numerator a bit, such that the derivative of the denominator appears in it.

This is needed here, so we add it and also subtract it.

And here, the derivative of the denominator appeared in the numerator, accompanied by some constant.

Now we decompose the fraction into the sum of two fractions:

The first integral is done:

We have to suffer a bit more with the other one.

We create a perfect square in the denominator.

Here, in the denominator, a perfect square appears. For the sake of simplicity, we will call the term behind it D.

We call these D, and voila!:

Now we should solve a problem.

We can easily integrate any rational function. All we have to do is decompose it into partial fractions, and then integrate the partial fractions using the previous methods.

There happens to be a problem here:

First let's check whether the degree of the numerator is less than that of the denominator. If this does not hold, then we will need to use polynomial long division.

The polynomial division is wonderful, we will check it out later, but luckily we don't need it now.

It is because the numerator is quadratic, and the denominator is cubic.

Let's start to decompose into partial fractions. For that, we need to factorize the denominator into linear and irreducible quadratic terms.

We can factor out x,

and this cannot be factorized further,

as it has a negative discriminant.

With this, factorization is done.

These are going to be the denominators of the partial fractions.

Now we have to figure out the numerators. Not the actual numerators just yet, but their parametric forms. Let's see what this means.

We are determined to decompose into partial fractions. And there are two types of partial fractions.

Type I partial fractions have a linear denominator.

Type II partial fractions have a non-factorable quadratic denominator.

During decomposition we always start with the denominators. The denominator of the first fraction is clearly linear, so it could only be a Type I partial fraction. So the numerator is some A.

On the other hand, the denominator of the second fraction is a quadratic polynomial, so this fraction is Type II, and thus its numerator is in the form of Ax+B.

Well, A is already in use, so let's say Bx+C will be the numerator.

And now we calculate what A, B and C are.

For this, we have to stare at them a bit.

We multiply by the denominators,

and then expand the parentheses.

And then we check how many x2 terms, x terms and constants are on the right side.

Because there are exactly the same on the left side as well.

We solve the system of equations.

The first fraction is done, the second will take a bit more time.

First we create the derivative of the denominator, then

We can easily integrate any rational function. All we have to do is decompose it into partial fractions, and then integrate the partial fractions using the previous methods.

There happens to be a problem here:

First let's check whether the degree of the numerator is less than that of the denominator. If this does not hold, then we will need to use polynomial long division.

The polynomial division is wonderful, we will check it out later, but luckily we don't need it now.

Let's factorize the denominator. Next, factor out x.

Then let's see whether the quadratic factor could be further factorized.

It seems, yes. If someone doesn't feel like doing such factorization, well, here is a splendid little formula:

where

Factorization is done. These are going to be the denominators of the partial fractions.

We figure out the numerators based on the denominators.

Since all three denominators are linear, all three factors are of Type I partial fractions, so the numerators are A, B and C.

Now we calculate what A, B and C are.

For this, we have to stare at them a bit.

We multiply by the denominators,

and then comes a trick.

Let's see what happens if we substitute zero for x.

Now let's try to compute the value of B.

For that, we should make these zero.

Finally, to compute C, we will make these zero.

If you did not like the trick, we could also expand the parentheses:

And then we check how many x2 terms, x terms and constants are on the right side.

Because there are exactly the same on the left side as well.

We solve the system of equations.

Here is another rational function:

Again, we need to factorize the denominator into linear and irreducible quadratic terms.

Let's see if this could be factorized.

Well, it seems, yes.

Now come the decomposing into partial fractions. As we can see, the denominator contains one of the linear factors twice. In such cases there is a small trick for the decomposition.

The denominator of one partial fraction is (2x+1), and the other's is (2x+1)2.

Again, we figure out the numerators based on the denominators.

Since all three denominators are linear, or a power of a linear term, all three factors are of Type I partial fractions, so the numerators are A, B and C.

Now let's see what A, B and C are.

We will use the tricky method shown in the previous slideshow.

First we zero these:

These cannot be zero at the same time, so finding A will be a bit more difficult.

Well, let's substitute 0 for x.

We could substitute 666, too but 0 is safer.

These are easy to integrate.


Summary problem for rational functions

Here is a summary problem where we can see all important steps.

Bad news. First, we need polynomial division, because the degree of the numerator must be less than the degree of the denominator.

This polynomial division is just like long division we learned in elementary school.

Like  25:7=3 and the remainder is 4.

In other words

Polynomial division is just like this.

quotient                                           remainder

Here comes the polynomial division:

Everything is OK so far.

But this is not the end, yet.

We multiply the quotient with the divisor,

and subtract that from the dividend.

And then we divide again, and repeat this until the degree of the dividend becomes less than that of the divisor.

Hey, it is less now, so we are done.

We integrate the first two members, and then we proceed to the fraction where the degree of the numerator is less than that of the denominator.

Again, we need to factorize the denominator into linear and irreducible quadratic factors. First, we factor out x2.

Then we check whether the remaining quadratic part could be factorized. The answer is no. It is because there is no real solution to the  equation.

On the other hand, x2 can be factorized.

Now come the decomposing into partial fractions.

If the denominator contains a square of some ax+b linear polynomial, then we decompose into partial fractions, such that

the denominator of one of the partial fractions is ax+b,

and the other is (ax+b)2.

Now we have to figure out the numerators. The denominator of the first fraction seems to be linear, so the numerator is some A.

The denominator of the second fraction is a square of a linear expression, so the numerator here is also some A, but since A is already used, let's call it B.

Finally, the denominator of the third fraction is a quadratic polynomial, so its numerator should be in the form of Ax+B, but since A and B are already used, Cx+D will do.

Now we calculate the values of A, B, C and D.

We multiply by the denominators.

Next, expand the parentheses.

And then we check how many x3 terms, x2 terms, x terms and constants are on the right side.

Because there are exactly the same on the left side as well.

The first two terms are very easy to integrate.

The third term becomes this:

The first term, as we wanted, f’/f, while the second term leads to arctangent.

And then it is done.

Finally, one more example:

First of all, we need to factorize the denominator into linear or irreducible quadratic factors. The factorization isn't trivial at all, because the denominator does not have a real root. The product form:

Then:

The factors in the denominator will be the denominators of the partial fractions. Since both factors are irreducible quadratic expressions, it seems we will get a sum of two Type II partial fractions:

Next, we move on to finding A, B, C, and D.

Multiplication:

and then conversion

finally the usual system of equations:

The solutions: , thus

We will integrate the two fractions separately.

The first fraction:

This will turn into a linear substitution of arctgx:

The second fraction, due to symmetry reasons:

The solution of the problem is the sum of the expressions found before:

Let's not forget though, that the method of integrating rational functions should be regarded as the last resort, and used only when all other methods failed to work. This latter problem was solved earlier, somewhat faster, using S4!


Integration of trigonometric terms 1.0 - The tangent half-angle substitution

INTEGRATION INVOLVING COMBINATIONS OF TRIGONOMETRIC FUNCTIONS

Integration of trigonometric terms is not easy. We will check out a few simpler tricks and the most important methods. Let's start with one of the most exciting methods.

One of the strangest cases of integration by substitution is when u= tan(x/2).

We use this if sinx and cosx is included in first-degree form in the fraction.

The main idea of the substitution is the following three formulae:

Now we will see some magic tricks.

Those who are not particularly fond of magic, can skip this part.

We start with a smaller trick:

Then, we use a more advanced trick.

At the end comes a finishing trick.

The hocus-pocus ends here, and we got this:

And, one more thing.

For the substitution, we need the derivative of the expressed x, too.

So, we need to express x first.

There is such a thing as:

This way, bye-bye tangent, we found x.

And we found the derivative, too. We know that

The good news is that we had to suffer through this previous ordeal only this one time.

From here, all we have to do is make a note of these, so we could refer to them when needed.

Here comes an exercise.

This method works in more complicated cases, too:

We do some decomposition

 and then it's done.

And that's all I have to say about that.


Integration of trigonometric terms 2.0

If either α or β is odd, then we hit the jackpot.

Let's see a specific example.

We will take the factor with the odd exponent, and turn it into a product of one linear factor and some quadratic factors.

Then comes a minor trick.

Finally we multiply, and an old friend appears:

It doesn't pose a problem if the exponent of the cosx is of higher degree.

We will again take the factor with the odd exponent, and turn it into a product of one linear factor and some quadratic factors.

Then, we will need this cubic formula:

Finally comes this one, again:

If both α and β are even, then this method won't work.

In this case we will use the so called linearization formulae.

Let's see one like this, too.

If we recall, there was this thing:

That's what we will use now.

The case of the snake biting its own tail

In these cases we use partial integration, and then do some magic tricks.

Let's see this one, for instance:

Here, the casting doesn't matter, we can assign the roles either way, the result will be the same.

Integral by parts, again.

And by this, we returned to the original problem.

If we continued the integration now, then two consecutive partial integrations would take us back to the original problem.

And we could continue that till the end of time. But that would be quite boring, so instead, here is a trick.

The main idea of the trick is to write an equation of what we found so far.

And then solve the equation.

Let's see one more:


Integration of trigonometric terms 3.0 - The case of the snake biting its own tail

If either α or β is odd, then we hit the jackpot.

Let's see a specific example.

We will take the factor with the odd exponent, and turn it into a product of one linear factor and some quadratic factors.

Then comes a minor trick.

Finally we multiply, and an old friend appears:

It doesn't pose a problem if the exponent of the cosx is of higher degree.

We will again take the factor with the odd exponent, and turn it into a product of one linear factor and some quadratic factors.

Then, we will need this cubic formula:

Finally comes this one, again:

If both α and β are even, then this method won't work.

In this case we will use the so called linearization formulae.

Let's see one like this, too.

If we recall, there was this thing:

That's what we will use now.

The case of the snake biting its own tail

In these cases we use partial integration, and then do some magic tricks.

Let's see this one, for instance:

Here, the casting doesn't matter, we can assign the roles either way, the result will be the same.

Integral by parts, again.

And by this, we returned to the original problem.

If we continued the integration now, then two consecutive partial integrations would take us back to the original problem.

And we could continue that till the end of time. But that would be quite boring, so instead, here is a trick.

The main idea of the trick is to write an equation of what we found so far.

And then solve the equation.

Let's see one more:


Summary problem for whole indefinite integration

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.


Problem I Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem I Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem I Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

And now, let’s see how to solve an integration problem.

Well, we have to ask ourselves a few questions.

Is there a thing like this in the integral:

If yes, then there are two possible cases.

There is a linear expression under the radical

There is a non-linear expression under the radical

In this case it is worth to try substitution:

In this case, it is usually a good idea to convert the radical thingy:

and then we can use S2

Is there  or  in the integral?

x is raised to the first power

x is raised to some power

This is presumably an S2

And this one is partial integration.

If SOMETHING is linear, then we should use partial integration.

If SOMETHING is not linear, then we need the S4 formula:

And now we can get to the problems.

We have to ask ourselves these questions before solving each of these problems, 

so that we could pick the appropriate formula.

Here is this one, for instance.

It contains a radical. Let’s see what we listed for radical cases like this:

It contains a logarithm. Let’s see what we listed for logarithmic cases like this:

It contains ex. Let’s see what we listed for ex cases like this:

And now let’s see what the difference is between the two problems.

At first sight, not much.

But let’s just see what we listed for ex cases like this:

Finally, we have these three.

There are radicals in all, but each must be solved in a different way.

Let’s see our notes:

Theoretically, this first one should be solved by substitution, but actually it is very simple.

Well, the second one is really for substitution, due to this x here.

.


Problem | Indefinite integral

Problem | Indefinite integral