If $\frac{1}{(x+1)(x^2+2x+2)}=\frac{A}{x+1}+\frac{Bx+C}{{(x+1)}^2+1}$then $A+B=$

Given$\frac{1}{(x+1)(x^2+2x+2)}=\frac{A}{x+1}+\frac{Bx+C}{{(x+1)}^2+1}$

$\frac{1}{(x+1)(x^2+2x+2)}=\frac{A({(x+1)}^2+1)+(Bx+C)(x+1)}{(x+1)(x^2+2x+2)}$

$1=A(x^2+2x+2)+(Bx+C)(x+1)$

Compare

Coefficient of $x^2\Rightarrow 0=A+B\to \left(1\right)$

Coefficient of$x\Rightarrow 0=2A+B+C\to \left(2\right)$

Constant term$\Rightarrow 1=2A+C\to \left(3\right)$

Solving (1), (2) and (3)

We get $A=1.B=-1,C=-1$

From (1) $A+B=0$