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?

Since the factors in the denominator can't be factored over in the real numbers further, the best you can do is$$\frac {s}{(s^2+2s+2)(s^2-2s+2)}=\frac {As+B}{s^2+2s+2}+\frac {Cs+D}{s^2-2s+2}$$
Multiply both sides by the denominator to clear the fractions to get the equation$$s=(As+B)(s^2-2s+2)+(Cs+D)(s^2+2s+2)$$
Now let replace any occurrence of $s^2$ with $2s-2$, simplify, and repeat the process with replacing $s^2$ with $-2s-2$.
