Answer
$\frac1 6 (t^4+1)^{\frac 3 2}+C$
Work Step by Step
$\int t^3\sqrt {t^4+1}dt$
$\frac 1 4 \int \sqrt u du$
$\frac 1 4 \int u^{1/2}=\frac 1 4 \int \frac {u^{3/2}}{3/2})$
$\frac 1 4 \times \frac 2 3 u^{3/2}$
$\frac1 6 (t^4+1)^{\frac 3 2}+C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
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 - Page 512: 18
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