Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - Review - Exercises - Page 577: 2


$$\ln \frac{3}{2}-\frac{1}{6} $$

Work Step by Step

Given $$\int_{1}^{2} \frac{x}{(x+1)^{2}} d x$$ Let $u=x+1\ \ \Rightarrow \ \ du=dx$ and at $ x=1\to u=2$, at $ x=2\to u=3$ \begin{align*} \int_{1}^{2} \frac{x}{(x+1)^{2}}d x&=\int_{2}^{3} \frac{ u-1 }{u^2} d u\\ &=\int_{2}^{3} \left( \frac{1 }{u}-u^{-2} \right)du\\ &= \ln u+\frac{1}{u}\bigg|_{2}^3\\ &=\ln \frac{3}{2} + \frac{1}{3}-\frac{1}{2} \\ &=\ln \frac{3}{2}-\frac{1}{6} \end{align*}
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