Answer
$$\frac{11}{3} \ln |x|-\frac{2}{x}-\frac{9}{2} \ln |x-1|+\frac{5}{6} \ln |x-3|+C$$
Work Step by Step
Given $$\int \frac{3 x+6}{x^{2}(x-1)(x-3)} d x$$
Since
\begin{aligned}
\frac{3 x+6}{x^{2}(x-1)(x-3)} &=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{(x-1)}+\frac{D}{(x-3)}\\
&=\frac{ A x(x-1)(x-3)+B(x-1)(x-3) +C x^{2}(x-3)+D x^{2}(x-1)}{x^{2}(x-1)(x-3)}\\
3 x+6&= A x(x-1)(x-3)+B(x-1)(x-3) +C x^{2}(x-3)+D x^{2}(x-1)
\
\end{aligned}
Then
\begin{align*}
\text{at } x&=1 \ \ \ \ \to C=\frac{-9}{2}\\
\text{at } x&=0\ \ \ \ \to B=2\\
\text{at } x&= 3\ \ \ \ \to D= \frac{5}{6}\\
\text{at } x&=-1\ \ \ \ \to A= \frac{11}{3}
\end{align*}
Hence
\begin{aligned}
\int \frac{(3 x+6) d x}{x^{2}(x-1)(x-3)}&=\int \frac{\frac{11}{3} d x}{x}+\int \frac{2 d x}{x^{2}}-\int \frac{\frac{9}{2} d x}{(x-1)}+\int \frac{\frac{5}{6} d x}{(x-3)}\\
&=\frac{11}{3} \ln |x|-\frac{2}{x}-\frac{9}{2} \ln |x-1|+\frac{5}{6} \ln |x-3|+C
\end{aligned}
