Answer
$$-\frac{1}{9} \ln |x-4|+\frac{1}{3} \frac{(x-4)^{-1}}{(-1)}+\frac{1}{9} \ln |x-1|+C$$
Work Step by Step
Given $$\int \frac{d x}{(x-4)^{2}(x-1)}$$
Since
\begin{aligned}
\frac{1}{(x-4)^{2}(x-1)} &=\frac{A}{(x-4)}+\frac{B}{(x-4)^{2}}+\frac{C}{(x-1)} \\
&= \frac{A(x-4)(x-1)+B(x-1)+C(x-4)^{2}}{(x-4)^{2}(x-1)} \\
1 &=A(x-4)(x-1)+B(x-1)+C(x-4)^{2}
\end{aligned}
Then
\begin{align*}
\text{at } x&=1\ \ \ \ \to C= \frac{1}{9} \\
\text{at } x&=4\ \ \ \ \to B=\frac{1}{3}\\
\text{at } x&=0\ \ \ \ \to A=\frac{-1}{9}
\end{align*}
Hence
\begin{aligned}
\int \frac{d x}{(x-4)^{2}(x-1)} &=-\frac{1}{9} \int \frac{d x}{(x-4)}+\frac{1}{3} \int(x-4)^{-2} d x+\frac{1}{9} \int \frac{d x}{(x-1)} \\
&=-\frac{1}{9} \ln |x-4|+\frac{1}{3} \frac{(x-4)^{-1}}{(-1)}+\frac{1}{9} \ln |x-1|+C
\end{aligned}
