Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.5 Strategy for Integration - 7.5 Exercises: 6

Answer

$\frac{1}{18}$

Work Step by Step

$\int_0^1\frac{x}{(2x+1)^3}dx$ Let $u=2x+1$. Then $du=2\ dx$, so $dx=\frac{1}{2}\ du$. Also, since $2x=u-1$, $x=\frac{u-1}{2}$. $\int_{2*0+1}^{2*1+1}\frac{\frac{u-1}{2}}{u^3}*\frac{1}{2}du$ $=\frac{1}{4}\int_1^3\frac{u-1}{u^3}du$ $=\frac{1}{4}\int_1^3(\frac{u}{u^3}-\frac{1}{u^3})du$ $=\frac{1}{4}\int_1^3(u^{-2}-u^{-3})du$ $=\frac{1}{4}(\frac{u^{-1}}{-1}-\frac{u^{-2}}{-2})|_1^3$ $=\frac{1}{4}(-\frac{1}{u}-\frac{1}{-2u^2})|_1^3$ $=\frac{1}{4}((-\frac{1}{3}-\frac{1}{-2*3^2})-(-\frac{1}{1}-\frac{1}{-2*1^2}))$ $=\frac{1}{4}((-\frac{1}{3}-\frac{1}{-18})-(-1-\frac{1}{-2}))$ $=\frac{1}{4}(-\frac{5}{18}-(-\frac{1}{2}))$ $=\frac{1}{4}*\frac{2}{9}$ $=\boxed{\frac{1}{18}}$
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