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?

Denote $F(x)$ this rational function. A priori, the decomposition into partial fractions over $\mathbf R$ has the form:
$$F(s)= \frac{As+B}{s^2+2s+2}+ \frac{Cs+D}{s^2-2s+2}.\tag{1}$$
We can determine the coefficients a little faster than with the basic method, if we observe that $F(s)$ is an odd function, so that
$$ F(-s)= \frac{-As+B}{s^2-2s+2} + \frac{-Cs+D}{s^2+2s+2}=-F(s)=-\frac{As+B}{s^2+2s+2}- \frac{Cs+D}{s^2-2s+2}$$
By identification, we obtain that $\;-As+B=-Cs-D$, whence
$$A=C,\quad B=-D.$$
Now, multiply both sides of $(1)$ by $s$ and let $s\to\infty$, we obtain the limits
$$0= A+C,\enspace\text{whence }\; A=C=0.$$
Next, set $x=1$:$$F(1)=\frac1{5\cdot 1}=\frac B5 +\frac D1=\frac{4D}5,\enspace\text{whence }\enspace D=\frac14,\;B=-\frac 14.$$
