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?!
 A: 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 $C$.
A: 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 $A =1, B = -1, C= -1$.
A: $$
\frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2}.
$$
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 $C$.
