Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.3 Integration By Substitution - Exercises Set 4.3: 21

Answer

$-\frac{1}{40(5x^4+2)^2}+C$

Work Step by Step

$\int\frac{x^3}{(5x^4+2)^3}dx$ Let $u=5x^4+2$. Then $du=20x^3dx$, so $x^3dx=\frac{du}{20}$. $=\int\frac{1}{u^3}*\frac{du}{20}$ $=\frac{1}{20}\int u^{-3}du$ $=\frac{1}{20}(\frac{u^{-2}}{-2}+C)$ $=-\frac{1}{40u^2}+C$ $=-\frac{1}{40(5x^4+2)^2}+C$
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