Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 95


$$\frac{1}{2}(\ln x)^2+c $$

Work Step by Step

Let $ u=\ln x $, then $ du= \frac{1}{x} dx $ and hence $$\int \frac{\ln x}{x}dx = \int udu=\frac{1}{2}u^2+c=\frac{1}{2}(\ln x)^2+c.$$
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