Answer
$=\frac{(x^4+3)^3}{12}+C$
Work Step by Step
$let$ $x^4+3=u$
$du=4x^3$ $dx$
$\int u^2$ $du$
$=\frac{1}{4}\int (x^4+3)^2$$4x^3$ $dx$
$=\frac{1}{4}(\frac{(x^4+3)^3}{3})+C$
$=\frac{(x^4+3)^3}{12}+C$
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 4 - Integration - 4.5 Exercises - Page 301: 9
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