Answer
$$
\int \frac{ d x}{9x-3}= \frac{1}{9}\ln |9x-3|+c.
$$
Work Step by Step
Le $ u=9x-3$ and hence $ du=9dx $, then we have
$$
\int \frac{ d x}{9x-3}=\frac{1}{9}\int \frac{ d u}{u}= \frac{1}{9}\ln |u|+c=\frac{1}{9}\ln |9x-3|+c.
$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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