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?
 A: 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
A: 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 $x=1$:$$F(1)=\frac1{5\cdot 1}=\frac B5 +\frac D1=\frac{4D}5,\enspace\text{whence }\enspace D=\frac14,\;B=-\frac 14.$$
A: Look for the decomposition in the form of: 
$$\dfrac{As+B}{(s-1)^2+1}+\dfrac{Cs+D}{(s+1)^2+1}.$$
Generally, when you have an irreducible of degree $n$ in the denominator, the numerator you need to guess need to have degree $n-1.$
A: You want numerators of degrees lower than your denominators. Let's note that $$\frac{s}{(s^2+2s+2)(s^2-2s+2)}=\frac{1}{4}(\frac{1}{s^2-2s+2}-\frac{1}{s^2+2s+2}).$$ Complex numbers are required to obtain partial fractions with linear denominators, viz. $$\frac{1}{s^2\pm 2s+2}=\frac{1}{(s\pm 1)^2+1}=\frac{i}{2}(\frac{1}{s\pm 1 +i}-\frac{1}{s\pm 1 -i}).$$
A: 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 $-2s-2$.
