Answer
$\frac{1}{4(1-x^2)^2}+C$
Work Step by Step
let $1-x^2=u$
$du=-2x$ $dx$
$\int \frac{x}{(1-x^2)^3}dx$
$=\frac{1}{2} \int \frac{1}{u^3}du$
$=-\frac{!}{2}(-\frac{1}{2}u^{-2})+C$
$=\frac{1}{4(1-x^2)^2}+C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 4 - Integration - 4.5 Exercises - Page 301: 17
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
