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: 100


$\frac{1}{2} (\ln|\ln x|)^2+c $

Work Step by Step

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