Integration of Rational Functions

Every rational function has an antiderivative that can be expressed in terms of some combination of rational functions, the natural logarithm, and arctangent.

Abstract

  • Basic case: we can use
  • We can also use partial fraction decomposition and then
  • Substitution is very useful, especially:
    • The condition being that the numerator and the differential match
    • If they don't match, make adjustments to the numerator and the break up the integral
  • For improper rational functions, use long division
  • Completing the square is very useful as well
  • Repeated quadratic denominators require a trig sub

Example - Basic

Example - Partial Fraction Decomp.

Evaluate

solution
We have to use partial fraction decomposition

When , , so
When , , so

For the last step, see log laws

Example - Substitution

Evaluate

solution
We know that $$\int \frac{1}{x^2 + 1} , d = \tan^{-1} (x) + C$$
we just have to deal with the . We can move the 4 out of the denominator, and then use substitution

Let , so

which is actually the formula described in integration rules

Example - Improper Rational Function

Evaluate

solution
We need to split up the improper rational function. Usually, this involves long division, but we can take a shortcut this time.

where the last step comes from the previous example.

Example - Completing the Square

Evaluate

solution
We can't use fraction decomposition, because the denominator is irreducible. But we can complete the square.

where the last step comes from the tangent formula

Substitution with Numerator != 1

Evaluate

solution
We use substitution
Let , then
Which is VERY GOOD, because is actually our numerator.

Example - Tricky Problem: Substitution with Numerator not Matching

Evaluate

solution
If we let , we find that . But that's not the numerator, so we have a troublemaker. Quid faciemus?

Well, we want in the numerator without changing the value of the fraction, so:

So,

How do we deal with the second term? Well, that's just the completing the square technique from a few examples ago.

We can proceed

Example - Repeated Quadratic

Evaluate

solution
We have a repeated quadratic in the denominator, so we need a trig sub.

Set , which means .

So,

(see half angle identities)

Now we want to return to , we use the half-angle formula for to obtain:

Since , we can create a triangle with as the base, as the height, and as the hypotenuse. Now, we can plug in: