Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Review - Exercises - Page 538: 15


$-\displaystyle \frac{1}{2}\ln|x|+\frac{3}{2}\ln|x+2|+C$

Work Step by Step

$ \displaystyle \frac{x-1}{x^{2}+2x}=\frac{x-1}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2}$ $\displaystyle \frac{x-1}{x^{2}+2x}=\frac{A(x+2)+Bx}{x(x+2)}=\frac{(A+B)x+2A}{x(x+2)}\Rightarrow\left\{\begin{array}{l} 2A=-1\\ A+B=1 \end{array}\right.$ $A=-\displaystyle \frac{1}{2},\quad B=\frac{3}{2}$ $\displaystyle \int\frac{x-1}{x^{2}+2x}=\int\left(\frac{-\frac{1}{2}}{x}+ \frac{\frac{3}{2}}{x+2} \right)dx $ $=-\displaystyle \frac{1}{2}\ln|x|+\frac{3}{2}\ln|x+2|+C$
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