Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 88


$$ \int \frac{ d x}{9x-3}= \frac{1}{9}\ln |9x-3|+c. $$

Work Step by Step

Le $ u=9x-3$ and hence $ du=9dx $, then we have $$ \int \frac{ d x}{9x-3}=\frac{1}{9}\int \frac{ d u}{u}= \frac{1}{9}\ln |u|+c=\frac{1}{9}\ln |9x-3|+c. $$
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