Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - Review - Exercises - Page 578: 15


$$\frac{-1}{2}\ln (x)+\frac{3}{2}\ln (x+2)+c $$

Work Step by Step

Given $$\int \frac{x-1}{x^2+2x }dx$$ Since \begin{align*} \frac{x-1}{x^2+2x }&=\frac{A}{x}+\frac{B}{x+2}\\ &=\frac{ A(x+2)+ B(x)}{x^2+2x}\\ x-1&= A(x+2)+ B(x) \end{align*} At $x=-2 \ \ \Rightarrow B=3/2$, and At $x=0 \ \ \Rightarrow A=-1/2$, then $$ \frac{x-1}{x^2+2x } =\frac{-1}{2x}+\frac{3}{2(x+2)}$$ Hence \begin{align*} \int \frac{x-1}{x^2+2x } dx&=\int \frac{-1}{2x}dx+\int\frac{3}{2(x+2)}dx\\ &=\frac{-1}{2}\ln (x)+\frac{3}{2}\ln (x+2)+c \end{align*}
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