Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.5 Strategy for Integration - 7.5 Exercises - Page 547: 11


$$\frac{1}{2} \left[ \sec^{-1}x+ \frac{\sqrt{x^2-1}}{x^2}\right]+c$$

Work Step by Step

Given $$ \int \frac{dx}{x^3\sqrt{x^2-1}}$$ Let $x=\sec u\ \ \Rightarrow \ dx=\sec u \tan udu$, then \begin{align*} \int \frac{dx}{x^3\sqrt{x^2-1}}&= \int \frac{ \sec u \tan udu}{\sec^3u\sqrt{\sec^2u-1}}\\ &= \int \frac{ \sec u \tan udu}{\sec^3u\tan u}\\ &= \int \frac{ du}{\sec^2u }\\ &= \int \cos^2u du\\ &=\frac{1}{2} \int(1+ \cos 2u)du\\ &=\frac{1}{2} [u+\frac{1}{2}\sin 2u]+c\\ &=\frac{1}{2} [u+ \sin u\cos u]+c\\ &=\frac{1}{2} \left[ \sec^{-1}x+ \frac{\sqrt{x^2-1}}{x^2}\right]+c \end{align*}
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.