Answer
= $x^{2}$$\sqrt {1+x^{3}}$
Work Step by Step
$\frac{d}{dx}$ [$\int_{{\,0}}^{{\,x}}$ $t^{2}$$\sqrt {1+t^{3}}$ $dt$ ] = $x^{2}$$\sqrt {1+x^{3}}$
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 4 - Integration - Review Exercises - Page 314: 59
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