Integration by Partial FractionsDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/