Inverse laplace 2/((s-1)^2+1)^2 
Find inverse laplace transform of $$\frac{2}{((s-1)^2+1)^2}$$

I tried to decompose the fraction using
$$\because (s-1)^2+1=s^2-2s+2$$
$$\rightarrow \frac{2}{((s-1)^2+1)^2}=\frac{As+B}{s^2-2s+2}+\frac{Cs+D}{(s^2-2s+2)^2}$$
yet I get D=2, which leads me back to the same exact equation, any help?
(I have also tried the $\frac{-d}{ds}(L{\{sint\}})$ propertie, but failed since there's no variable s in the numerator)

While there is a defined inverse Laplace transform, the integral is often difficult.  Usually, it is easier to take the transforms we know and noodle with them until we find a fit.
$L\{e^t\cos t\} = \frac {s-1}{(s-1)^2+1}\\
L\{e^t\sin t\} = \frac 1{(s-1)^2+1} = \frac {s^2-2s +2}{((s-1)^2+1)^2}\\
L\{te^t\cos t\} = -\frac {d}{ds} \frac {(s-1)}{(s-1)^2+1} = \frac {s^2 - 2s}{((s-1)^2+1)^2}\\
L\{te^t\sin t\} = -\frac {d}{ds} \frac 1{(s-1)^2+1} = \frac {2s-2}{((s-1)^2+1)^2}$
Some combination of $(s^2 - 2s + 2), (s^2-2s),(2s-2) = 2$
$e^t \sin t - te^t \cos t$ 
