Answer
$ \frac{1}{2}(\tan ^{-1} t)^2+c.$
Work Step by Step
We have
$$
\int \frac{\tan ^{-1} t d t}{1+t^{2}}=\int \tan ^{-1} t \ d (\tan ^{-1} t)\\
= \frac{1}{2}(\tan ^{-1} t)^2+c.
$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 56
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