Given 3 x - 2 ( x + 1 ) 2 ( x + 3 ) = A x + 1 + B ( x + 1 ) 2 + C x + 3 , then 4 A + 2 B + 4 C

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

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

$3x-2=A(x^2+3+4x)+B(x+3)+C(x^2+1+2x)$

$\Sigma \text{ of coefficients of }{x}^2=0 \Rightarrow A+C=0$

$\Sigma \text{ of coefficients of x}=3 \Rightarrow 4A+B+2C=3$

$\Sigma \text{of coefficients of constants}=0 \Rightarrow 3A+3B+C=-2$

Equating the three equations, we get

$A=-C, B=\frac{-2}{5}$

$4A+2B+4C=-5$