Given$$y''+2y'+2y=3\sin x+\cos x$$
Transform to image region $$Y(s)(s^2+2s+2)=\frac{3}{s^2+1}+\frac{s}{s^2+1}-s-2$$ $$Y(s)((s^2+2s+1)+1)=\frac{3}{s^2+1}+\frac{s}{s^2+1}-s-2$$ $$Y(s)((s+1)^2+1)=\frac{3}{s^2+1}+\frac{s}{s^2+1}-s-2$$ $$Y(s)=\frac{3}{(s^2+1)((s+1)^2+1)}+\frac{s}{(s^2+1)((s+1)^2+1)}-\frac{s}{((s+1)^2+1)}-\frac{2}{((s+1)^2+1)}$$
Now I need the inverse of $$\frac{1}{((s+1)^2+1)}$$ to continue. Can't find it anywhere
Hint: $\mathscr{L^{-1}}\left(\frac{b}{(s-a)^2+b^2}\right) = e^{at}\sin{(bt)}$; in your case, $a=-1$ and $b=1$. You can prove it by the integral definition of Laplace transformation. More info here.
Another hint:, you can also find the inverse of $\frac{s}{((s+1)^2+1)}$ by smart usage of the $\mathscr{L^{-1}}$ of $e^{at}\cos{(bt)}$ in the link above. (That is, $s=s+0=s+(1-1)$)