Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 275: 13


$$ \int \frac{t^3}{(4-2t^4)^{11}} dx =\frac{1}{80}\frac{1}{(4-2t^4)^{10}}+c. $$

Work Step by Step

Since $ u=4-2t^4$, then $ du=-8t^3dt $ and hence, $$ \int \frac{t^3}{(4-2t^4)^{11}} dx=-\frac{1}{8}\int \frac{1}{u^{11}} d u=-\frac{1}{8}\int u^{-11} d u\\ =-\frac{1}{8} \frac{-1}{10}u^{-10} +c=\frac{1}{80}\frac{1}{(4-2t^4)^{10}}+c. $$
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.