University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 12


$\dfrac{[\ln (\ln x)]^2}{2} +c$

Work Step by Step

Solve $\int\dfrac{\ln (\ln x)}{x \ln x}dx$ Let us $p= \ln (\ln x)$ and $dp=\dfrac{1}{x \ln x}dx$ This implies $\int\dfrac{\ln (\ln x)}{x \ln x}dx= \int p dp$ $= \dfrac{p^2}{2} +c$ $= \dfrac{[\ln (\ln x)]^2}{2} +c$ Hence, $\int\dfrac{\ln (\ln x)}{x \ln x}dx=\dfrac{[\ln (\ln x)]^2}{2} +c$
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