Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.3 Integration By Substitution - Exercises Set 4.3 - Page 286: 18



Work Step by Step

Use the substitution $u=4-5x^2$. Then $du=-10xdx$, so $xdx=-\frac{1}{10}du$. $\int\frac{x}{\sqrt{4-5x^2}}dx$ $=\int\frac{1}{\sqrt{u}}*(-\frac{1}{10})du$ $=\int-\frac{1}{10}u^{-\frac{1}{2}}du$ $=-\frac{1}{10}*\frac{u^\frac{1}{2}}{\frac{1}{2}}+C$ $=-\frac{\sqrt{u}}{5}+C$ $=-\frac{\sqrt{4-5x^2}}{5}+C$
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.