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: 72


$$\frac{1}{\ln 5}\sec^{-1}5^x+c$$

Work Step by Step

Let $u=5^x$, then $du=5^x\ln 5 dx$, now we have $$\int \frac{dx}{\sqrt{5^{2x}-1}}=\frac{1}{\ln 5}\int \frac{du}{u\sqrt{u^2-1}} \\ =\frac{1}{\ln 5}\sec^{-1}u+c=\frac{1}{\ln 5}\sec^{-1}5^x+c.$$
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.