Answer
$=\frac{(5x^2+4)^4}{40}+C$
Work Step by Step
$let$ $5x^2+4=u$
$du=10x$ $dx$
$\int u^3$ $du$
$=\frac{1}{10}\int (5x^2+4)^3$$10x$ $dx$
$=\frac{1}{10}(\frac{(5x^2+4)^4}{4})+C$
$=\frac{(5x^2+4)^4}{40}+C$
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: 12
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