Calculus 10th Edition

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

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises: 18

Answer

$\frac1 6 (t^4+1)^{\frac 3 2}+C$

Work Step by Step

$\int t^3\sqrt {t^4+1}dt$ $\frac 1 4 \int \sqrt u du$ $\frac 1 4 \int u^{1/2}=\frac 1 4 \int \frac {u^{3/2}}{3/2})$ $\frac 1 4 \times \frac 2 3 u^{3/2}$ $\frac1 6 (t^4+1)^{\frac 3 2}+C$
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