# Find the Inverse Laplace transform of: F(s)=(1)/(s(s^2+8s+4))

$F\left(s\right)=\frac{1}{s\left({s}^{2}+8s+4\right)}$
After completing the square, I obtained
$F\left(s\right)=\frac{1}{s\left(\left(s+4{\right)}^{2}-12\right)}$
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ruinsraidy4
Using partial fraction decomposition gives
$F\left(s\right)=\frac{1}{s\left({s}^{2}+8s+4\right)}=\frac{1}{4}\left(\frac{1}{s}-\frac{s}{{s}^{2}+8s+4}-\frac{8}{{s}^{2}+8s+4}\right)$
$\therefore F\left(s\right)=\frac{1}{4}\left(\frac{1}{s}-\frac{s}{\left(s+4{\right)}^{2}-12}-\frac{8}{\left(s+4{\right)}^{2}-12}\right)$
Then in order to use
$\mathcal{L}\left\{{e}^{at}\right\}=\frac{1}{s-a}$
$\mathcal{L}\left\{{e}^{at}\mathrm{sinh}\left(bt\right)\right\}=\frac{b}{\left(s-a{\right)}^{2}-{b}^{2}}$
$\mathcal{L}\left\{{e}^{at}\mathrm{cosh}\left(bt\right)\right\}=\frac{s-a}{\left(s-a{\right)}^{2}-{b}^{2}}$
we can rewrite F(s) as
$F\left(s\right)=\frac{1}{4}\left(\frac{1}{s}-\frac{s+4}{\left(s+4{\right)}^{2}-12}-\frac{4}{\left(s+4{\right)}^{2}-12}\right)$
$=\mathcal{L}\left\{\frac{1}{4}\left(1-{e}^{-4t}\mathrm{cosh}\left(\sqrt{12}t\right)-\frac{4}{\sqrt{12}}{e}^{-4t}\mathrm{sinh}\left(\sqrt{12}t\right)\right)\right\}$