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 - Page 301: 14



Work Step by Step

$\int t^3 (2t^4+3)^{\frac{1}{2}}dt$ let $2t^4+3=u$ $8t^3dt=du$ $t^3dt=\frac{1}{3}du$ $\int t^3 (2t^4+3)^{\frac{1}{2}}dt$ $=\frac{1}{8} \int u^{\frac{1}]{2}}$ $=\frac{1}{12}u^{\frac{3}{2}}+C$ $=\frac{(2t^4+3)^{\frac{3}{2}}}{12}+C$
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