Answer
$=-\frac{(6-x^3)^6}{18}+C$
Work Step by Step
$let$ $6-x^3=u$
$du=-3x^2$ $dx$
$\int u^5$ $du$
$=-\frac{1}{3}\int (6-x^3)^5$$-3x^2$ $dx$
$=-\frac{1}{3}(\frac{(6-x^3)^6}{6})+C$
$=-\frac{(6-x^3)^6}{18}+C$
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: 10
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