Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises - Page 334: 25



Work Step by Step

let u = $x-1$ du=dx $\int\frac{2x}{(x-1)^2}dx=2\int\frac{x}{(x-1)^2}dx$ for $\int\frac{x}{(x-1)^2}dx$, write $x$ as $x-1+1$ $=2\int(\frac{1}{x-1}+\frac{1}{(x-1)^2})dx$ $=2(\int\frac{1}{x-1}dx+\int\frac{!}{(x-1)^2}dx)$ $=2(ln(|x-1|)-\frac{1}{x-1})+C$
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