Find: $\mathcal{L}^{-1} \left(\frac{2s}{(s^2 + 4)^2}\right)$

Question

I tried to compute the inverse Laplace transform: $\mathcal{L}^{-1} \left(\frac{2s}{(s^2 + 4)^2}\right)$ By decomposing the fraction to the form of $\frac{As+B}{s^2+4}+\frac{Cs+D}{(s^2+4)^2}$ , but this lead nowhere. Any help, thank you all.

Answer 1

$$\mathcal{L}\{\sin{(at)}\}=\frac{a}{s^2+a^2}$$ $$\begin{align} \mathcal{L}\{t\sin{(at)}\} &=-\frac{\mathrm{d}}{\mathrm{d}s}\mathcal{L}\{\sin{(at)}\}\\ &=\frac{2as}{(s^2+a^2)^2}\\ \end{align}$$ $$\therefore\mathcal{L}^{-1}\left\{\frac{2s}{(s^2+4)^2}\right\}=\frac12t\sin{(2t)}$$