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: 11



Work Step by Step

Use the substitution $u=4x-3$. This means $du=4dx$, so $dx=\frac{1}{4} du$. $\int (4x-3)^9dx$ $=\int u^9*\frac{1}{4} du$ $=\int \frac{1}{4}u^9 du$ $=\frac{1}{4}*\frac{u^{10}}{10}+C$ $=\frac{u^{10}}{40}+C$ $=\frac{(4x-3)^{10}}{40}+C$
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