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: 35


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

Work Step by Step

$$A=\int \frac{\ln x}{x^2}dx$$ We set $u=\ln x$ and $dv=\frac{dx}{x^2}$ That means $du=\frac{dx}{x}$ and $v=-\frac{1}{x}$ Following the formula $\int udv=uv-\int vdu$, we have $$A=-\frac{\ln x}{x}-\int-\frac{1}{x}\frac{dx}{x}$$ $$A=-\frac{\ln x}{x}+\int\frac{1}{x^2}dx$$ $$A=-\frac{\ln x}{x}-\frac{1}{x}+C$$ $$A=\frac{-\ln x-1}{x}+C$$
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