Answer
$\frac{(x^3-1)^5}{15}+C$
Work Step by Step
$\int x^2(x^3-1)^4dx$
let
$x^3-1=u$
$3x^2dx=du$
$x^2dx=\frac{1}{3}du$
$\int x^2(x^3-1)^4dx$
$=\frac{1}{3}\int u^4du$
$=\frac{1}{15}u^5+C$
$=\frac{(x^3-1)^5}{15}+C$
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: 11
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