Answer
$\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
$$\int \frac{\ln|\ln x|}{ x\ln x}dx = \int udu\\
=\frac{1}{2} u^2+c
=\frac{1}{2} (\ln|\ln x|)^2+c.$$
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: 100
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