Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.5 - Strategy for Integration - 7.5 Exercises - Page 508: 45


$$\int x^{5}e^{-x^{3}}dx=\frac{-1}{3e^{x^{3}}}(x^{3}+1)+C$$

Work Step by Step

$$let\ t=x^{3},\,dt=3x^{2}dx$$ $$\int x^{5}e^{-x^{3}}dx=\frac{1}{3}\int te^{-t}dt$$ $$=-\frac{1}{3}te^{-t}+\frac{1}{3}\int e^{-t}dt=-\frac{1}{3}te^{-t}-\frac{1}{3}e^{-t}+C$$ $$=\frac{-1}{3e^{x^{3}}}(x^{3}+1)+C$$
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