# Integration of Rational Functions Using Partial Fractions

 Today: 7.4: Integration of rational functions and Supp. 4: Partial fraction expansion Next: 7.7: Approximate integration

Our goal today is to compute integrals of the form

by decomposing . This is called partial fraction expansion.

Theorem 5.5.1 (Fundamental Theorem of Algebra over the Real Numbers)   A real polynomial of degree can be factored as a constant times a product of linear factors and irreducible quadratic factors .

Note that , where , are complex conjugates.

Types of rational functions . To do a partial fraction expansion, first make sure using long division. Then there are four possible situation, each of increasing generality (and difficulty):

1. is a product of distinct linear factors;

2. is a product of linear factors, some of which are repeated;

3. is a product of distinct irreducible quadratic factors, along with linear factors some of which may be repeated; and,

4. is has repeated irreducible quadratic factors, along with possibly some linear factors which may be repeated.

The general partial fraction expansion theorem is beyond the scope of this course. However, you might find the following special case and its proof interesting.

Theorem 5.5.2   Suppose , and are polynomials that are relatively prime (have no factor in common). Then there exists polynomials and such that

Proof. Since and are relatively prime, using the Euclidean algorithm (long division), we can find polynomials and such that

Dividing both sides by and multiplying by yields

which completes the proof.

Example 5.5.3   Compute

First do long division. Get quotient of and remainder of . This means that

Since we have distinct linear factors, we know that we can write

for real numbers . A clever way to find is to substitute appropriate values in, as follows. We have

Setting on both sides we have (taking a limit):

Likewise, we have

Thus

Example 5.5.4   Compute the partial fraction expansion of . By the partial fraction theorem, there are constants such that

Note that there's no possible way this could work without the term, since otherwise the common denominator would be . We have

This method will not get us ! For example,

While true this is useless.

Instead, we use that we know and , and evaluate at another value of , say 0.

so . Thus finally,

 constant

Example 5.5.5   Let's compute . Notice that is a factor, since is a root. We have

There exist constants such that

Then

You could find by factoring the quadratic over the complex numbers and getting complex number answers. Instead, we evaluate at a couple of values. For example, at we get

so . Next, use to get .

so

Finally,

It remains to compute

First, complete the square to get

Let , so and . Then

Finally, we put it all together and get

Discuss second quiz problem.

Problem: Compute using complex exponentials. The answer is

Here's how to get it.

Simplify the inside part requires some imagination:

William Stein 2006-03-15