find the inverse of Laplace transform frac{3}{(s+2)^2}-frac{2s+6}{(s^2+4)}

Step 1
Given Laplace transform is,
$\frac{3}{(s+2{)}^2}-\frac{2s+6}{(s^2+4)}$
Step 2
Taking inverse of Laplace transformation as,
$L^{-1}\{\frac{3}{(s+2{)}^2}-\frac{2s+6}{(s^2+4)}\}=L^{-1}\left\{\frac{3}{(s+2{)}^2}\right\}-L^{-1}\left\{\frac{2s+6}{(s^2+4)}\right\}$
$=3L^{-1}\left\{\frac{1}{(s+2{)}^2}\right\}-2L^{-1}\left\{\frac{s}{s^2+4}\right\}-6L^{-1}\left\{\frac{1}{s^2+4}\right\}$
Step 3
Some formulas of inverse of Laplace transformation are
1) $L^{-1}\{F(s)\}=f(t)\text{ then }{L}^{-1}\{F(s-a)\}=e^{at}f(t)$
$\Rightarrow L^{-1}\left\{\frac{1}{(s+2{)}^2}\right\}=e^{-2t}{L}^{-1}\left\{\frac{1}{s^2}\right\}=e^{-2t}t$
2)$L^{-1}\left\{\frac{a}{s^2+a^2}\right\}=\sin at$
$\Rightarrow L^{-1}\left\{\frac{1}{s^2+4}\right\}=\frac{1}{2}{L}^{-1}\left\{\frac{2}{s^2+4}\right\}=\frac{(\sin 2t)}{2}$
3)$L^{-1}\{sF(s)\}=f^{\prime }(t)+f(0)$
Then $L^{-1}\left\{\frac{s}{s^2+4}\right\}=\frac{1}{2}{L}^{-1}\left\{s\frac{2}{s^2+4}\right\}=\frac{1}{2}((\sin 2t{)}^{\prime }+\sin (0))=\cos 2t$
Step 4
Then, consider equation (1),
$3L^{-1}\left\{\frac{1}{(s+2{)}^2}\right\}-2L^{-1}\left\{\frac{s}{s^2+4}\right\}-6L^{-1}\left\{\frac{1}{s^2+4}\right\}=3L^{-1}\left\{\frac{1}{(s+2{)}^2}\right\}-2L^{-1}\left\{\frac{s}{s^2+4}\right\}-\frac{6}{2}{L}^{-1}\left\{\frac{2}{s^2+4}\right\}=3e^{-2t}t-2\cos 2t-3\sin 2t$
Step 5
Therefore, the inverse of the given Laplace transformation is,
$L^{-1}\{\frac{3}{(s+2{)}^2}-\frac{2s+6}{s^2+4}\}=3e^{-2t}t-2\cos 2t-3\sin 2t$