University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 429: 72



Work Step by Step

$$A=\int\tan^{-1}xdx$$ Take $y=f^{-1}(x)=\tan^{-1}x$ So $x=f(y)=\tan y$ Applying the formula $\int f^{-1}(x)dx=xf^{-1}x-\int f(y)dy$, we have $$A=x\tan^{-1}x-\int\tan ydy$$ $$A=x\tan^{-1}x-\ln|\sec y|+C$$ $$A=x\tan^{-1}x+\ln|\sec(\tan^{-1}x)|+C$$
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