Answer
$$\frac{1}{3}(\ln x)^3+c.$$
Work Step by Step
Let $ u=\ln x $, then $ du= \frac{1}{x} dx $ and hence
$$\int \frac{(\ln x)^2 }{x}dx = \int u^2du=\frac{1}{3}u^3+c\\
=\frac{1}{3}(\ln x)^3+c.$$
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: 97
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