Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises: 86

Answer

$\frac{1}{3}e^{x^3}+C$

Work Step by Step

Let $u=x^3$. Then $du=3x^2 dx$, and $\frac{1}{3}du=x^2dx$. $\int x^2 e^{x^3}dx$ $=\int e^{x^3}x^2 dx$ $=\int e^u*\frac{1}{3}du$ $=\frac{1}{3}e^u+C$ $=\frac{1}{3}e^{x^3}+C$
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