Given, $\frac{3 x}{(x-2)(x+1)}$ can be written as,


$\begin{aligned}
& 3 x=A(x+1)+B(x-2) \\
& \text { At } x=2, \\
& \therefore \quad A=\frac{3(2)}{2+1} \Rightarrow A=\frac{3(2)}{3}=A=2 \\
& \text { At } x=-1, B=\frac{3(-1)}{-1-2} \\
& B=\frac{-3}{-3} \Rightarrow B=1
\end{aligned}$
Substitute the values of $A$ and $B$ in Eq. (i), we get
$\begin{aligned}
& \frac{3 x}{(x-2)(x+1)}=\frac{2}{x-2}+\frac{1}{x+1} \\
& =\frac{2}{-2\left(1-\frac{x}{2}\right)}+\frac{1}{(x+1)}=-\left(1-\frac{x}{2}\right)^{-1}+(1+x)^{-1} \\
& =-\left[1+\frac{x}{2}+\left(\frac{x}{2}\right)^2+\ldots+\left(\frac{x}{2}\right)^5+\ldots .\right]
\end{aligned}$
$$
+\left(1-x+x^2-x^3+x^4-x^5+\ldots .\right)
$$
$\therefore \quad$ Coefficient of $x^5=-\left(\frac{1}{2}\right)^5-1=\frac{-1}{32}-1=-\frac{33}{32}$