Next: Bibliography Up: 17. Antidifferentiation Techniques Previous: 17.6 Substitution in Integrals   Index

# 17.7 Rational Functions

In this section we present a few rules for finding antiderivatives of simple rational functions.

To antidifferentiate where is a polynomial, make the substitution .

17.56   Example. To find .

Let . Then so , and

To find where and is a polynomial of degree less than .

We will find numbers and such that

 (17.57)

Suppose (17.57) were valid. If we multiply both sides by we get

Now take the limit as goes to to get

The reason I took a limit here, instead of saying now for we get '' is that is not in the domain of the function we are considering. Similarly

and if we take the limit as goes to , we get

Thus,
 (17.58)

I have now shown that if there are numbers and such that (17.57) holds, then (17.58) holds. Since I have not shown that such numbers exist, I will verify directly that (17.58) is valid. Write . Then

17.59   Example. To find .

Let .

Then

so

and

so

Hence

In this example I did not use formula (17.58), because I find it easier to remember the procedure than the general formula. I do not need to check my answer, because my proof of (17.58) shows that the procedure always works. (In practice, I usually do check the result because I am likely to make an arithmetic error.)

To find where is a polynomial of degree , and does not factor as a product of two first degree polynomials.

Complete the square to write

Then , since if then we have factored , and if we can write , and then

and again we get a factorization of . Since , we can write for some , and

Now

Make the substitution to get an antiderivative of the form

The last antiderivative can be found by a trigonometric substitution.

17.60   Example. To find :

Let

Let , so and . Then

Now let , so , and . Then

Hence,

To find where is a polynomial of degree .

First use long division to write

where is a polynomial, and is a polynomial of degree . Then use one of the methods already discussed.

17.61   Example. To find . By using long division, we get

Hence

17.62   Example. In exercise 17.7, you showed that is an antiderivative for . The function in that exercise appeared magically with no motivation. I will now derive the formula, using standard methods:

Now let . Then , and

Suppose . Then

and if we take the limit of both sides as we get . Also

and if we take the limit as , we get . Thus

Now

so

and thus

17.63   Exercise. Criticize the following argument:

I want to find . Suppose

Then

If we take the limit of both sides as , we get . Also

and if we take the limit of both sides as , we get . Thus

Hence,

17.64   Exercise. Find the following antiderivatives:

a)

b)

c)

d)

e)

f)

g)
A

17.65   Exercise. Find the following antiderivatives:
a)
.
b)
.
c)
.
d)
.
e)
.
f)
.
g)
.
h)
.
i)
Choose a number , and find .
j)
Choose a number , and find .
k)
.
l)
.
m)
.
n)
.
o)
.
p)
.

Next: Bibliography Up: 17. Antidifferentiation Techniques Previous: 17.6 Substitution in Integrals   Index
Ray Mayer 2007-09-07