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: 99


$\frac{1}{4}\ln|8x-2|+c $

Work Step by Step

Let $ u=\ln |8x-2|$, then $ du= \frac{8}{8x-2} dx=\frac{4}{4x-1} dx $ and hence $$\int \frac{1}{(4x-1)\ln (8x-2)}dx =\frac{1}{4} \int \frac{1}{u}du\\ =\frac{1}{4}\ln |u|+c =\frac{1}{4}\ln|8x-2|+c.$$
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