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: 20


$-4^{-1}(4x^3-9)^{-2} +C$

Work Step by Step

let $4x^3-9=u$ $du=12x^2$ $dx$ $\int \frac{6x^2}{(4x^3-9)^3}dx$ $=2 \int \frac{du}{u^3} $ $=2 (-\frac{1}{2} u^{-2})+C$ $=-4^{-1}(4x^3-9)^{-2} +C$
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