Partial fraction with a constant as numerator

You set it up in the usual way as $$\begin{align*}\frac1{(x+1)(x^2+2x+2)}&=\frac{A}{x+1}+\frac{Bx+C}{x^2+2x+2}\\ &=\frac{A(x^2+2x+2)+(Bx+C)(x+1)}{(x+1)(x^2+2x+2)}\;, \end{align*}$$

so that $$A(x^2+2x+2)+(Bx+C)(x+1)=1\;.$$

Now multiply out the lefthand side to get $$(A+B)x^2+(2A+B+C)x+(2A+C)=1$$

and equate coefficients of $x^2,x$, and $1$ to get (in that order):

$$\left\{\begin{align*} &A+B=0\\ &2A+B+C=0\\ &2A+C=1 \end{align*}\right.$$

Then solve the system for $A,B$, and $C$.

You want to find $A, B,$ and $C$ such that

$$\frac{1}{(x+1)(x^2 + 2x + 2)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 + 2x + 2} $$ That is such that $$\begin{align}0x^2 + 0x + 1 &= A(x^2 + 2x + 2) + (x+1)(Bx+c)\\ &= (A+B)x^2 + (2A+B+C)x + 2A + C. \end{align}$$ So you get three equations $$\begin{align} 0 &= A + B \\ 0 &= 2A + B + C \\ 1 &= 2A + C. \end{align}$$ Solving this I get $A =1, B = -1, C= -1$.