Answer
$$3 \ln |x-1|-2 \ln |x+1|-\frac{5}{(x+1)}+C$$
Work Step by Step
Given $$\int \frac{\left(x^{2}+11 x\right) d x}{(x-1)(x+1)^{2}}$$
Since
\begin{align*}
\frac{\left(x^{2}+11 x\right) }{(x-1)(x+1)^{2}}&=\frac{A}{(x-1)}+\frac{B}{(x+1)}+\frac{C}{(x+1)^{2}}\\
&=\frac{A(x+1)^{2}+B \cdot(x-1)(x+1)+C(x-1)}{(x-1)(x+1)^{2}}\\
x^{2}+11 x&=A(x+1)^{2}+B \cdot(x-1)(x+1)+C(x-1)
\end{align*}
Then
\begin{align*}
\text{at } x&=1 \ \ \ \ \to A=3 \\
\text{at } x&= -1\ \ \ \ \to C=5\\
\text{at } x&= 0\ \ \ \ \to B=-2
\end{align*}
Hence
\begin{aligned}
\int \frac{\left(x^{2}+11 x\right) d x}{(x-1)(x+1)^{2}} &=\int \frac{3 d x}{(x-1)}-\int \frac{2 d x}{(x+1)}+\int \frac{5 d x}{(x+1)^{2}} \\
&=3 \int \frac{d x}{(x-1)}-2 \int \frac{d x}{(x+1)}+5 \int(x+1)^{-2} d x\\
&=3 \ln |x-1|-2 \ln |x+1|-\frac{5}{(x+1)}+C
\end{aligned}
