Answer
$\frac{1}{48}(x^3+9)^{16}+c$
Work Step by Step
Let $u=x^3+9$, then $du=3x^2dx$. Hence, we have
$$\int x^2(x^3+9)^{15}dx=\frac{1}{3}\int u^{15}du\\
=\frac{1}{3}\frac{1}{16}u^{16}+c=\frac{1}{48}(x^3+9)^{16}+c.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 40
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