Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.4 Integrals Involving Hyperbolic and Inverse Hyperbolic Functions - Exercises - Page 415: 17


$$\cosh ^{-1} x+C $$

Work Step by Step

Since $$\frac{d}{d x} \cosh ^{-1} x=\frac{1}{\sqrt{x^{2}-1}} $$ Then $$ \int \frac{1}{\sqrt{x^{2}-1}} d x=\cosh ^{-1} x+C $$
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.