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