Summary problem for rational functions | mathXplain
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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, Polynomial division.

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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!

# Summary problem for rational functions

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Calculus 1 episode
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