Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Integration by Partial Fraction Decomposition

Date: 06/24/99 at 05:20:12
From: Robert
Subject: Integral problem

I am having problems doing this kind of problem:

     integral of dx/(x^4+x^2)

I tried factoring an x^2 out, but this doesn't seem to help.  Also,

     integral of xdx/(x^2+2x+1)

I tried doing this by parts,

     u= x, du = dx   and    dv = 1/((x+1)^2+1), v = arctan(x+1)

but this doesn't seem to work either. HELP!


Date: 06/24/99 at 08:13:07
From: Doctor Jerry
Subject: Re: Integral problem

Hi Robert,

What you need is "partial fraction decomposition."

     1/(x^4+x^2) = 1/((x^2+1)(x^2)).

Using partial fraction principles, we know that this fraction came 
from fractions of the form

     (A*x+B)/(x^2+1) + C/x + D/x^2,

where A, B, C, and D are unknowns, to be determined. In this 
particular case, C = 0, D = 1, A = 0, and B = -1. That is,

     1/(x^4+x^2) = 1/x^2 - 1/(1+x^2).

Now the integration is easy.

For the other,

     x/(x^2+2x+1) = x/(x+1)^2.

Here the partial fraction decomposition must have the form

     A/(x+1) + B/(x+1)^2.

In this case, A = 1 and B = -1. Now the integration is easy.

How do you find A and B? Set

     A/(x+1) + B/(x+1)^2 = x/(x+1)^2

Clear fractions by multiplying by (x+1)^2:

                A(x+1)+B = x.

This must hold for all x. Let x take on two nice values to obtain two 
equations in two unknowns. Here, take x = -1 and x = 0:

                       B = -1

                     A+B = 0

So, B = -1 and A = 1.

- Doctor Jerry, The Math Forum
Associated Topics:
College Calculus

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994-2013 The Math Forum