Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 49

Answer

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

Work Step by Step

Let $u=x^2-4=>du=2x dx$ and $\frac{1}{2}du=xdx$ thus $\int \frac{x}{(x^2-4)^3}dx$ =$\int (x^2-4)^{-3}xdx$ =$\int u^{-3}\frac{1}{2}du=\frac{1}{2}\int u^{-3}du$ =$\frac{-1}{4}u^{-2}+C$ =$\frac{-1}{4}(x^2-4)^{-2}+C$
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