Contents of this Calculus 1 episode:

Integrating rational functions, Partial fractions, First type partial fractions, Second type partial fractions, Decompose it into partial fractions, Factorization, Linear denominator, Non-factorable quadratic denominator, f’/f and arctan forms.

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.

Calculus 1 episode