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: 67


$\frac{1}{10}(2y+3)^{5/2}-\frac{1}{2} (2y+3)^{3/2}+c.$

Work Step by Step

Let $u=2y+3$, $du=2dy$. We have, \begin{align*} \int y(2y+3)^{1/2}dy&=\frac{1}{4}\int (u-3)u^{1/2}du\\ &=\frac{1}{4}\int u^{3/2}-3u^{1/2}du\\ &=\frac{1}{4}\frac{1}{5/2}u^{5/2}-\frac{1}{4}\frac{3}{3/2}u^{3/2}+c\\ &=\frac{1}{10}(2y+3)^{5/2}-\frac{1}{2} (2y+3)^{3/2}+c. \end{align*}
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