Hello, jeng!
You didn't tell us exactly
where your difficulty is . . .
\(\displaystyle \L\int \frac{5x\,+\,1}{x(x\,-\,3)^2}\,dx\)
The set-up? \(\displaystyle \L\;\frac{5x\,+\,1}{x(x\,-\,3)^2}\;= \;\frac{A}{x}\,+\,\frac{B}{x\,-\,3}\,+\,\frac{C}{(x\,-\,3)^2}\)
Clearing denominators? \(\displaystyle \;5x\,+\,1\;=\;A(x\,-\,3)^2\,+\,B(x(x\,-\,3) \,+\,Cx\)
Solving for \(\displaystyle A,\,B,\,C\) ?
Let \(\displaystyle x\,=\,0:\;1\:=\:9A\;\;\Rightarrow\;\;A\,=\,\frac{1}{9}\)
Let \(\displaystyle x\,=\,3:\;\;16\,=\,3C\;\;\Rightarrow\;\;C\,=\,\frac{16}{3}\)
Let \(\displaystyle x\,=\,1\;\;6\:=\:4A\,\,2B\,+\,C\;\;\Rightarrow\;\;B\,=\,-\frac{1}{9}\)
Hence: \(\displaystyle \L\,\frac{5x\,+\,1}{x(x\,-\,3)^2}\;=\;\frac{\frac{1}{9}}{x}\,+\,\frac{-\frac{1}{9}}{x\,-\,3} \,+\,\frac{\frac{16}{3}}{(x\,-\,3)^2}\)
Integrating ?
\(\displaystyle \L\;\;\frac{1}{9}\int\frac{dx}{x}\,-\,\frac{1}{9}\int\frac{dx}{x\,-\,3} \,+ \,\frac{16}{3}\int\)\(\displaystyle (x\,-\,3)^{-2}\,dx\)
\(\displaystyle \L\;=\;\frac{1}{9}\ln|x|\,-\,\frac{1}{9}\ln|x\,-\,3|\,-\,\frac{16}{3}(x\,-\,3)^{-1}\,+\,C\)
\(\displaystyle \L\;=\;\frac{1}{9}\ln\left|\frac{x}{x\,-\,3}\right|\,-\,\frac{16}{3(x\,-\,3)} \,+\,C\)
