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