Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 70


$$\frac{1}{2} (\tan^{-1}x)^2+c.$$

Work Step by Step

Let $u= \tan^{-1}x$, then $du =\frac{dx}{x^2+1}$. Now, we have $$\int \frac{\tan^{-1}x dx}{x^2+1}=\int udu\\ =\frac{1}{2} u^2+c=\frac{1}{2} (\tan^{-1}x)^2+c.$$
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