Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.5 Exercises: 16

Answer

$\frac{2}{9}(u^3+2)^{\frac{3}{2}}+C$

Work Step by Step

let $u^3+2=p$ $dp=3u^2 du$ $\int u^2(u^3+2)^{\frac{1}{2}}du$ $=\frac{1}{3} \int p^{\frac{1}{2}} dp$ $=\frac{1}{3}(\frac{2}{3}p^{\frac{3}{2}})+C$ $=\frac{2}{9}(u^3+2)^{\frac{3}{2}}+C$
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