# determine L^{-1}(F) F(s)=frac{4}{(s+2)^3}

determine ${L}^{-1}\left(F\right)$
$F\left(s\right)=\frac{4}{\left(s+2{\right)}^{3}}$
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Step 1
Given Data
The function is $F\left(s\right)=\frac{4}{\left(s+2{\right)}^{3}}$
The general inverse transform rule is
${L}^{-1}\left(F\left(s\right)\right)=f\left(t\right)$
${L}^{-1}\left(F\left(s-a\right)\right)={e}^{at}f\left(t\right)$
Step 2
The inverse Laplace transform of function $F\left(s\right)=\frac{4}{\left(s+2{\right)}^{3}}$ using the above expression is,
${L}^{-1}\left(F\left(s\right)\right)={L}^{-1}\left(\frac{4}{\left(s+2{\right)}^{3}}\right)$
$={e}^{-2t}{L}^{-1}\left(\frac{4}{\left(s+2-2{\right)}^{3}}\right)$
$={e}^{-2t}{L}^{-1}\left(\frac{4}{{s}^{3}}\right)$
$={e}^{-2t}\cdot 2{t}^{2}$
Hence the inverse Laplace transform of function $F\left(s\right)=\frac{4}{\left(s+2{\right)}^{3}}$ is ${e}^{-2t}\cdot 2{t}^{2}$