How to find inverse Laplace transform? Please I am trying to solve inverse Laplace transform of $ \frac 6 {(s^2+4)^2}$
If anyone has some idea to help me please share it with me.
I have been trying to do like this but it's not working,
$ \frac 2 {s^2+4} \frac 3 {s^2+4} $ 
but then second fraction is a problem
 A: Hint:
Use convolution theorem:

$$\mathcal{L}\{f(t)*g(t)\}=F(s)G(s)$$
  where $F(s)=\mathcal{L}\{f(t)\}$ and $G(s)=\mathcal{L}\{g(t)\}$. The convolution of two functions $f$ and $g$ is defined as
  $$(f*g)(t)=\int_0^{t}f(x)g(t-x)dx$$


Let $F(s)=\frac{2}{s^2+4}$, $f(t)=\sin (2t)$, $G(s)=\frac{3}{s^2+4}$, and $g(t)=\frac{3}{2}\sin (2t)$. 
It follows that 
\begin{align*}
\mathcal{L}^{-1}\{\frac{2}{s^2+4} \frac{3}{s^2+4}\}&=\mathcal{L}^{-1}\{F(s)G(s)\}\\
&=\int_{0}^tf(x)g(x-t)dx\\
&=\frac{3}{2}\int_0^t\sin(2t)\sin[2(x-t)]dx
\end{align*}
A: Since:
$$ \mathcal{L}^{-1}\left(\frac{1}{s-k}\right) = e^{kx},\qquad  \mathcal{L}^{-1}\left(\frac{1}{(s-k)^2}\right) = x\,e^{kx}\tag{1}$$
and by partial fraction decomposition/the residue theorem:
$$ \frac{6}{(s^2+4)^2} = \frac{-\frac{3i}{16}}{s-2i}+\frac{\frac{3i}{16}}{s+2i}+\frac{-\frac{3}{8}}{(s+2i)^2}+\frac{-\frac{3}{8}}{(s-2i)^2}\tag{2}$$
we have:
$$ \mathcal{L}^{-1}\left(\frac{6}{(s^2+4)^2}\right) = \frac{3}{8}\left(\sin(2x)-2x\cos(2x)\right).\tag{3}$$
A: Another quick way to approach this problem that circumvents convolution integrals and partial fraction expansions is to use the following properties of the Laplace Transform.
$$
\begin{align}
\mathscr{L}^{-1}\left(\frac{a}{s^2+a^2}\right)(t)&=\sin at\,u(t) \tag 1\\\\
\mathscr{L}^{-1}\left(\frac{dF(s)}{ds}\right)(t)&=-tf(t)u(t) \tag 2\\\\
\mathscr{L}^{-1}\left(\frac{G(s)}{s}\right)(t)&=\int_0^t g(t')dt' \tag 3
\end{align}$$
Then, noting that $\frac{6}{(s^2+4)^2}=-\frac{3}{s}\frac{d}{ds}\left(\frac{1}{s^2+4}\right)$ and invoking $(1)-(3)$ gives 
$$\begin{align}
\mathscr{L}^{-1}\left(\frac{6}{(s^2+4)^2}\right)(t)&=-3\mathscr{L}^{-1}\left(\frac1s\frac{d}{ds}\left(\frac{1}{s^2+4}\right)\right)(t)\\\\
&=-3\int_{0}^t\left(-t'\left(\frac12 \sin(2t')\right)\right)\,dt\\\\
&=\frac38 \sin (2t)-\frac34 t\cos (2t)
\end{align}$$
as expected!
A: Define the complex-valued function:
$$F(s)=\dfrac {6 e^{st}} {(s+2j)^2 (s-2j)^2}$$
and the compute the residues:
$$\mathrm{Res}(F(s),2j)=\dfrac 1 {1!} \lim_{s \to 2j} \left[\dfrac d {ds} (s-2j)^2F(s) \right]=-\frac {3j} {16}e^{2jt}-\frac {3t} {8}e^{2jt}$$
$$\mathrm{Res}(F(s),-2j)=\dfrac 1 {1!} \lim_{s \to -2j} \left[\dfrac d {ds} (s+2j)^2F(s) \right]=\frac {3j} {16}e^{-2jt}-\frac {3t} {8}e^{-2jt}$$
The inverse Laplace is then obtained as
$$f(t)=\frac 1 {2 \pi j} \oint_{C}{\dfrac {6 e^{st}} {(s+2j)^2 (s-2j)^2}ds}=\mathrm{Res}(F(s),2j)+\mathrm{Res}(F(s),-2j)$$
After some straight forward mathematical manipulations the desired final result can be obtained. 
