University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 7 - Section 7.3 - Hyperbolic Functions - Exercises - Page 418: 40


See the result below.

Work Step by Step

We need to verify both sides of the expression. In order to do this, we will differentiate both sides. $\dfrac{d}{dx} ( \int x tanh^{-1} x dx)=\dfrac{d}{dx} ( x \tanh^{-1} x +\dfrac{1}{2} \ln (1-x^2) + C)$ or, $ tanh^{-1} x = x(\dfrac{1}{1-x^2})+(\tanh^{-1} x) (1)+\dfrac{1}{2}(\dfrac{-2x}{1-x^2})+(0)$ or, $tanh^{-1} x = tanh^{-1} x $ Hence, the result has been verified.
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