Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Review - True-False Quiz - Page 553: 1

Answer

True

Work Step by Step

True. Let $u=\tan^{-1}(x)$ and $dv=dx$. Then, $du=\frac{1}{x^2+1}dx$ and $v=x$. Using the Integration by Parts, $\int \tan^{-1}x dx=\tan^{-1}x\cdot x-\int x\cdot \frac{1}{x^2+1}dx$ $=x\tan^{-1}x-\int \frac{x}{x^2+1}dx$ $=x\tan^{-1}x-\frac{1}{2}\ln |x^2+1|+C$
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