University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 49


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

Work Step by Step

$$A=\int \frac{x}{(x^2-4)^3}dx$$ We set $u=x^2-4$ Then $$du=2xdx$$ $$xdx=\frac{1}{2}du$$ Therefore, $$A=\frac{1}{2}\int\frac{1}{u^3}du=\frac{1}{2}\int u^{-3}du$$ $$A=\frac{1}{2}\times\frac{u^{-2}}{-2}+C$$ $$A=-\frac{u^{-2}}{4}+C=-\frac{1}{4u^2}+C$$ $$A=-\frac{1}{4(x^2-4)^2}+C$$
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