Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 20


$$-\frac{1}{x}(\ln x+1)+c.$$

Work Step by Step

We do the integration by parts; we choose $u=\ln x$ and $dv=\frac{1}{x^2}dx$. Then, $du=\frac{dx}{x}$, $v=-\frac{1}{x}$, hence we have $$\int \frac{\ln x}{x^2}dx=\int udv=uv-\int vdu\\ =-\frac{\ln x}{x}+\int x^{-2}dx\\ =-\frac{\ln x}{x}-x^{-1}+c \\ =-\frac{1}{x}(\ln x+1)+c.$$
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