$$
\begin{aligned}
& \frac{3 x+2}{(x+1)\left(2 x^2+3\right)}=\frac{A}{x+1}+\frac{B x+C}{2 x^2+3} \\
& (3 x+2)=A\left(2 x^2+3\right)+(x+1)(B x+C)
\end{aligned}
$$
when
$$
\begin{aligned}
x & =-1 \\
5 A & =-1 \quad \Rightarrow A=-\frac{1}{5}
\end{aligned}
$$
When
$$
\begin{aligned}
& x=0 \quad \Rightarrow 3 A+C=2 \\
& C=2+\frac{3}{5} \Rightarrow \frac{13}{5}
\end{aligned}
$$
When
$$
x=1
$$
$$
\begin{aligned}
5 A+2 B+2 C & =5 \\
-1+2 B+\frac{26}{5} & =5 \Rightarrow 2 B=6-\frac{26}{5} \\
2 B & =\frac{4}{5} \Rightarrow B=\frac{2}{5}
\end{aligned}
$$


$$
\therefore \quad A-B+C=-\frac{1}{5}-\frac{2}{5}+\frac{13}{5}=\frac{10}{5}=2
$$