Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 430: 40


See the proof below.

Work Step by Step

Verify that both sides of the equation are equal: $\dfrac{d}{dx} ( \int x tanh^{-1} x dx)=\dfrac{d}{dx} ( x \tanh^{-1} x +\dfrac{1}{2} \ln (1-x^2) + C)$ or, $ tanh^{-1} x = x(\dfrac{1}{1-x^2})+(\tanh^{-1} x) (1)+\dfrac{1}{2}(\dfrac{-2x}{1-x^2})+(0)$ Hence, $tanh^{-1} x = tanh^{-1} x $ $(\bf{Verified})$
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