University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 64


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

Work Step by Step

$$A=\int \frac{\sqrt{\tan^{-1}x}}{1+x^2}dx$$ We set $u=\tan^{-1}x$ Then $$du=\frac{1}{1+x^2}dx$$ Therefore, $$A=\int \sqrt udu=\int u^{1/2}du$$ $$A=\frac{u^{3/2}}{\frac{3}{2}}+C=\frac{2u^{3/2}}{3}+C$$ $$A=\frac{2}{3}(\tan^{-1}x)^{3/2}+C$$
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