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