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: 8

Answer

$\frac{3}{4}(3-4x^2)^{\frac{4}{3}}+C$

Work Step by Step

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