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