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: 14

Answer

$\frac{(2t^4+3)^{\frac{3}{2}}}{12}+C$

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.