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: 106


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

Work Step by Step

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