Partial fraction with a constant as numerator I am trying to express this as partial fraction:
$$\frac{1}{(x+1)(x^2+2x+2)}$$
I have a similar exaple that has $5x$ as numerator, it is easy to understand. I do not know what to do with 1 in the numerator, how to solve it?!

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 $C$.
