Integration by Partial Fractions

Date: 11/14/96 at 07:08:35
From: Anonymous
Subject: Calculus question

Hi,

How do you integrate f(x) = x^2/[(x^4) + 1]?

Thanks for your help.

Date: 11/14/96 at 10:58:41
From: Doctor Anthony
Subject: Re: Calculus question

The integral can be solved by first factoring (x^4 + 1) into the 
quadratic factors (x^2 - sqrt(2)x +1)(x^2 + sqrt(2)x + 1) and then 
using partial fractions to express the fraction as:
 
(1/2sqrt(2))[x/(x^2 - sqrt(2)x +1)] - (1/2sqrt(2))[x/(x^2+sqrt(2)x + 
1)]

You are now going to have to make use of an integration formula that 
can be found in a table.  Specifically, the one that for b^2 < 4ac:

INT[x/(ax^2 + bx + c)]dx = (1/2a)log(ax^2 + bx + c) - (b/2a){2/
(sqrt(4ac-b^2))arctan[(2ax + b)/sqrt(4ac-b^2)]}

This means that the answer to your integral should be:

1/(4sqrt(2))log(x^2-xsqrt(2)+1) + 1/(2sqrt(2))arctan[(2x-sqrt(2))/
sqrt(2)] - 
1/(4sqrt(2))log(x^2+xsqrt(2)+1) + 1/(2sqrt(2))arctan[(2x+sqrt(2))/
sqrt(2)]

-Doctors Anthony, Ken, and Rachel,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/