Calculus (3rd Edition)

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

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 58



Work Step by Step

Let $u=t+8$, then $du=dt$. Now, we have \begin{align*} \int t^2(t+8)^{1/2} dt&=\int (u-8)^2u^{1/2} du\\ &=\int u^{5/2} -16u^{3/2}+64 u^{1/2}du\\ &=\frac{1}{7/2}u^{7/2}-\frac{16}{5/2}u^{5/2}+\frac{64}{3/2}u^{3/2}+c\\ &=\frac{2}{7}(t+8)^{7/2}-\frac{32}{5}(t+8)^{5/2}+\frac{128}{3}(t+8)^{3/2}+c. \end{align*}
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