Answer
$$\frac{1}{2} (\tan^{-1}x)^2+c.$$
Work Step by Step
Let $u= \tan^{-1}x$, then $du =\frac{dx}{x^2+1}$. Now, we have
$$\int \frac{\tan^{-1}x dx}{x^2+1}=\int udu\\
=\frac{1}{2} u^2+c=\frac{1}{2} (\tan^{-1}x)^2+c.$$
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 - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 70
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