Solution:

$\frac{3 x+2}{(x+1)^{2}(x-1)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x-1}$

$\therefore 3 x+2=A(x-1)(x+1)+B(x-1)+C(x+1)^{2}$

$put x = -1$

$\therefore -3 + 2 = B(-1-1)$

$B = \frac{1}{2}$

$put x = 1$

$\therefore 3 + 2 = C (1+1^{2})$

$C = \frac{5}{4}$

$Put x = 0,B = \frac{1}{2}, C = \frac{5}{4}$

$\therefore 2=A(0-1)(0+1)+\frac{1}{2}(0-1)+\frac{5}{4}(0+1)^{2}$

$A = -\frac{5}{4}$

$\therefore \frac{3 x+2}{(x+1)^{2}(x-1)}=\frac{-\frac{5}{4}}{x+1}+\frac{\frac{1}{2}}{(x+1)^{2}}+\frac{\frac{5}{4}}{x-1}$