Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 96



Work Step by Step

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