Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 387: 68


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

Work Step by Step

Let $ u=x^3$, then $ du=3x^2dx $. Now, we have $$\int x^2e^{x^2}dx =\frac{1}{3}\int e^udu =\frac{1}{3}e^u+c=\frac{1}{3}e^{x^3}+c.$$
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