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

Answer

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

Work Step by Step

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