Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.5 Exercises: 17

Answer

$\frac{1}{4(1-x^2)^2}+C$

Work Step by Step

let $1-x^2=u$ $du=-2x$ $dx$ $\int \frac{x}{(1-x^2)^3}dx$ $=\frac{1}{2} \int \frac{1}{u^3}du$ $=-\frac{!}{2}(-\frac{1}{2}u^{-2})+C$ $=\frac{1}{4(1-x^2)^2}+C$
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