Partial fraction of $\frac{s}{(s^2+2s+2)(s^2-2s+2)}$ I am trying to find the partial fraction of: 
$$\frac{s}{(s^2+2s+2)(s^2-2s+2)}$$
I started off with:
$$\frac{A(s^2-2s+2)}{s^2+2s+2} + \frac{B(s^2+2s+2)}{s^2-2s+2}$$
After that I get the following equations:
$A+B = 0$; $-2A+2B =1$
Giving: $A=-B$, and secondly $4B=1$, Hence:
$A=-\frac{1}{4}$ and $B=\frac{1}{4}$
But is this correct?

Let $$\frac{s}{(s^2+2s+2)(s^2-2s+2)}=\frac{A(s^2+2s+2)+B(s^2-2s+2)}{(s^2+2s+2)(s^2-2s+2)}$$
The solutions to $$s^2+2s+2=0\space\text{are}\space s=-1\pm i$$
Using $s=-1+i$, we find $-1+i=B(4-4i)$, yielding $B=-\frac 14$. 
Moreover, the solutions to$$s^2-2s+2=0\space\text{are}\space s=1\pm i$$
Using $s=1+i$, we get $1+i=A(4+4i)$, yielding $A=\frac 14$. Hence we can say that:
$$\frac{s}{(s^2+2s+2)(s^2-2s+2)}=\frac{1}{4(s^2-2s+2)}-\frac{1}{4(s^2+2s+2)}$$ 
Both graphed here
