University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 343: 72


$$\int \frac{(\tan^{-1}x)^2dx}{1+x^2}=\frac{(\tan^{-1}x)^3}{3}+C$$

Work Step by Step

$$A=\int \frac{(\tan^{-1}x)^2dx}{1+x^2}$$ We set $a=\tan^{-1}x$, which means $$da=\frac{dx}{1+x^2}$$ Therefore, $$A=\int a^2da$$ $$A=\frac{a^3}{3}+C$$ $$A=\frac{(\tan^{-1}x)^3}{3}+C$$
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