Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.1 Exercises - Page 251: 10



Work Step by Step

Given: $\int \frac{1}{(3x)^{2}}dx$ Rewrite: $=\int \frac{1}{9x^{2}}dx$ $=\frac{1}{9}\int x^{-2}dx$ Integrate: $=\frac{1}{9}(\frac{x^{-1}}{-1})+C$ Simplify: $=-\frac{1}{9x}+C$
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