University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 59


$$\int\frac{(\ln x)^{-3}}{x}dx=-\frac{1}{2(\ln x)^2}+C$$

Work Step by Step

$$A=\int\frac{(\ln x)^{-3}}{x}dx$$ We set $a=\ln x$, which means $$da=\frac{1}{x}dx$$ Therefore, $$A=\int a^{-3}da$$ $$A=\frac{a^{-2}}{-2}+C$$ $$A=-\frac{1}{2a^2}+C$$ $$A=-\frac{1}{2(\ln x)^2}+C$$
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