Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.5 Exercises: 11

Answer

$\frac{(x^3-1)^5}{15}+C$

Work Step by Step

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