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 - Page 301: 8



Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.