Calculus (3rd Edition)

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



Work Step by Step

Let $u=x^3+9$, then $du=3x^2dx$. Hence, we have $$\int x^2(x^3+9)^{15}dx=\frac{1}{3}\int u^{15}du\\ =\frac{1}{3}\frac{1}{16}u^{16}+c=\frac{1}{48}(x^3+9)^{16}+c.$$
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