Answer
$\frac{3}{4}(3-4x^2)^{\frac{4}{3}}+C$
Work Step by Step
let $3-4x^2=y$
$du=-8x$ $dx$
$\int(3-4x^2)^{\frac{1}{3}}(-8x)dx$
$=\int u^{\frac{1}{3}}du$
$=\frac{3}{4}u^{\frac{4}{3}}+C$
$=\frac{3}{4}(3-4x^2)^{\frac{4}{3}}+C$
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: 8
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