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 - Page 286: 19



Work Step by Step

Use the substitution $u=1-2x$. Then $du=-2dx$, so $dx=-\frac{1}{2}du$. $\int \frac{6}{(1-2x)^3}dx$ $=\int \frac{6}{u^3}*(-\frac{1}{2})du$ $=\int -\frac{3}{u^3}du$ $=\int -3u^{-3}du$ $=\frac{-3u^{-2}}{-2}+C$ $=\frac{3}{2u^2}+C$ $=\frac{3}{2(1-2x)^2}+C$
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