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: 38

Answer

See the explanation below.

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.