$\begin{aligned}
& \frac{3 x+1}{(x+2)(x-1)^2}=\frac{A}{x+2}+\frac{B}{(x-1)}+\frac{C}{(x-1)^2} \\
& \Rightarrow 3 x+1=A(x-1)^2+B(x-1)(x+2)+C(x+2) \\
& \Rightarrow 0 \cdot x^2+3 x+1=(A+B) x^2 \\
& \quad+(-2 A+B+C) x+(A-2 B+2 C)
\end{aligned}$
On comparing, we get
\begin{array}{l|l|l}
\hlineA+B=0 & -2 A+B+C=3 & A-2 B+2 C=1 \\
\hline\Rightarrow A=-B \ldots (i) & \Rightarrow 3 B+C=3 \ldots (ii) & \Rightarrow-3 B+2 C=1 \ldots (iii) \\
\hline
\end{array}
Adding Eqs. (ii) and (iii),
$3 C=4 \Rightarrow C=4 / 3$
From Eq. (ii), we get
$\begin{array}{rlrl}
3 B+\frac{4}{3} & =3 \Rightarrow 3 B=3-\frac{4}{3}=\frac{5}{3} \\
\Rightarrow & B & =5 / 9
\end{array}$
From Eq. (i), we get
$A=-B=\frac{-5}{9}$
Hence, $\frac{3 x+1}{(x+2)(x-1)^2}=\frac{-5}{9}\left(\frac{1}{x+2}\right)+\frac{5}{9}\left(\frac{1}{x-1}\right)$
$+\frac{4}{3}\left(\frac{1}{(x-1)^2}\right)$