$\because \quad \frac{3 x+2}{(x+1)\left(2 x^2+3\right)}=\frac{A}{x+1}+\frac{B x+C}{2 x^2+3}$
$\Rightarrow \quad 3 x+2=A\left(2 x^2+3\right)+(B x+C)(x+1)$
On putting $x+1=0$
$3(-1)+2=A\left[2(-1)^2+3\right]$
$\Rightarrow \quad-3+2=A(5)$
$\Rightarrow \quad A=-\frac{1}{5}$
Now on comparing the coefficients of like power of $x$
Coefficient of $x^2$
$0=2 A+B$
$\Rightarrow \quad B=\frac{2}{5}$
and coefficient of $x$
$3=B+C$
$\Rightarrow \quad C=3-\frac{2}{5}=\frac{13}{5}$
$\therefore \quad A+C-B=-\frac{1}{5}+\frac{13}{5}-\frac{2}{5}$
$=\frac{13-2-1}{5}=\frac{10}{5}=2$