Answer
$$\frac{1}{2}(\ln x)^2+c $$
Work Step by Step
Let $ u=\ln x $, then $ du= \frac{1}{x} dx $ and hence
$$\int \frac{\ln x}{x}dx = \int udu=\frac{1}{2}u^2+c=\frac{1}{2}(\ln x)^2+c.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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