University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 38


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

Work Step by Step

$$A=\int x^5e^{x^3}dx=\int x^3e^{x^3}(x^2dx)$$ We set $a=x^3$. We then have $$da=3x^2dx$$ $$x^2dx=\frac{1}{3}da$$ Therefore, $$A=\frac{1}{3}\int ae^ada$$ Next, take $u=a$ and $dv=e^ada$ We would have $du=da$ and $v=e^a$ Applying the formula $\int udv=uv-\int vdu$, we have $$A=\frac{1}{3}(ae^a-\int e^ada)$$ $$A=\frac{1}{3}(ae^a-e^a)+C$$ $$A=\frac{1}{3}(x^3e^{x^3}-e^{x^3})+C$$
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