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