University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 34


$$\int \frac{1}{x(\ln x)^2}dx=-\frac{1}{\ln x}+C$$

Work Step by Step

$$A=\int \frac{1}{x(\ln x)^2}dx=\int\frac{1}{(\ln x)^2}\frac{dx}{x}$$ Set $a=\ln x$ We then have $$da=\frac{dx}{x}$$ Therefore, $$A=\int\frac{1}{a^2}da$$ $$A=-\frac{1}{a}+C$$ $$A=-\frac{1}{\ln x}+C$$
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