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: 71


$$\int \frac{dy}{\sqrt{\tan^{-1}y}(1+y^2)}=2\sqrt{\tan^{-1}y}+C$$

Work Step by Step

$$A=\int \frac{dy}{\sqrt{\tan^{-1}y}(1+y^2)}$$ We set $a=\sqrt{\tan^{-1}y}$, which means $$da=\frac{(\tan^{-1}y)'dy}{2\sqrt{\tan^{-1}y}}=\frac{dy}{2\sqrt{\tan^{-1}y}(1+y^2)}$$ $$\frac{dy}{\sqrt{\tan^{-1}y}(1+y^2)}=2da$$ Therefore, $$A=\int 2da$$ $$A=2a+C$$ $$A=2\sqrt{\tan^{-1}y}+C$$
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