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 402: 46


$$\int\frac{dx}{x(\log_{8}x)^2}=-\frac{(\ln8)^2}{\ln x}+C$$

Work Step by Step

$$A=\int\frac{dx}{x(\log_{8}x)^2}=\int\frac{dx}{\frac{x(\ln x)^2}{(\ln8)^2}}$$ $$A=(\ln8)^2\int\frac{dx}{x(\ln x)^2}$$ We set $u=\ln x$, which means $$du=\frac{dx}{x}$$ Therefore, $$A=(\ln8)^2\int\frac{1}{u^2}du$$ $$A=(\ln8)^2\times\Big(-\frac{1}{u}\Big)+C$$ $$A=-\frac{(\ln8)^2}{\ln x}+C$$
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