<s>[INST] 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?!
 [/INST] $$
\frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2}.
$$
Then you need to find $A$, $B$, and $C$.
The polynomial $x^2+2x+2$ factors as $(x+1+i)(x+1-i)$, where $i$ is a square root of $-1$.  You could go on to write
$$
\frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2} = \frac{A}{x+1} + \frac{D}{x+1+i} + \frac{E}{x+1-i},
$$
and the numbers $D$ and $E$ might not be real.
Later edit in response to a comment:
$$
\frac{1}{(x+1)(x^2+2x+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2}.
$$
There are several methods for finding $A$, $B$, and $C$.
Either of two methods begins by multiplying both sides by the denominator, getting
$$
1 = A(x^2+2x+2) + (Bx+C)(x+1).
$$
In one method, you can make the second term vanish by setting $x=-1$:
$$
1 = A\Big((-1)^2+2(-1)+2\Big) + \Big(\text{the second term, which is now }0\Big).
$$
Solving this for $A$ gives $A=1$, and then you write
$$
1 = 1(x^2+2x+2) + (Bx+C)(x+1).
$$
Now you can let $x=1$ in order to make the term involving $B$ vanish:
$$
1 = 1(0^2+2\cdot0 + 2) + C(0+1).
$$
This gives $C=-1$.  Then we have
$$
1 = (x^2+2x+2) + (Bx-1)(x+1).
$$
We can no longer make anything vanish without making $B$ disappear, so let $x=\text{some number that won't make the arithmetic too messy}$.  If we let $x=1$, we get
$$
1 = (1^2+2\cdot1+2) + (B2-1)(1+1).
$$
This gives us $B=-1/2$.
Another method is this: We had
$$
1 = A(x^2+2x+2) + (Bx+C)(x+1).
$$
Now multiply this out and collect like terms, where "like" means they are coefficients of the same powers of $x$:
$$
0x^2+0x^1 = (A+B)x^2 + (2A+B+C)x + (2A+C).
$$
Then equate coefficients of common powers of $x$:
$$
\begin{align}
A+B & = 0 \\
2A+B+C & = 0 \\
2A+C & = 1
\end{align}
$$
Then solve that system for $A$, $B$, and $C$.
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