Use the table of laplace transform to find the inverse Laplace transform of F(s)=\frac{2+s(s+1)}{s(s^2-s-6)}

Use the table of laplace transform to find the inverse Laplace transform of $F\left(s\right)=\frac{2+s\left(s+1\right)}{s\left({s}^{2}-s-6\right)}$
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Step 1
solution: We need to find the inverse laplace transform of $F\left(s\right)=\frac{2+s\left(s+1\right)}{s\left({s}^{2}-s-6\right)}$
Step 2
For this
$F\left(s\right)=\frac{2+s\left(s+1\right)}{s\left({s}^{2}-s-6\right)}$
$=\frac{2}{s\left({s}^{2}-s-6\right)}+\frac{s+1}{\left({s}^{2}-s-6\right)}$
$F\left(s\right)=\frac{2}{s\left(s-3\right)\left(s+2\right)}+\frac{s+1}{\left(s-3\right)\left(s+2\right)}$
using partial fraction
$F\left(s\right)=-\frac{1}{3s}+\frac{2}{15}\frac{1}{s-3}+\frac{1}{5}\frac{1}{\left(s+2\right)}+\frac{4}{5}\frac{1}{\left(s-3\right)}+\frac{1}{5\left(s+2\right)}$
Taking inverse laplace transform is
$f\left(t\right)={L}^{-1}\left\{F\left(s\right)\right\}=-\frac{1}{3}{L}^{-1}\left\{\frac{1}{s}\right\}+\frac{2}{15}{L}^{-1}\left\{\frac{1}{s-3}\right\}+\frac{1}{5}{L}^{-1}\left\{\frac{1}{s+2}\right\}+\frac{4}{5}{L}^{-1}\left\{\frac{1}{s-3}\right\}+\frac{1}{5}{L}^{-1}\left\{\frac{1}{s+2}\right\}$
$f\left(t\right)=-\frac{1}{3}+\frac{2{e}^{3t}}{15}+\frac{{e}^{-2t}}{5}+\frac{4}{5}{e}^{3t}+\frac{{e}^{-2t}}{5}$
$f\left(t\right)=\frac{14}{15}{e}^{3t}+\frac{2}{5}{e}^{-2t}-\frac{1}{3}$
Hence, inverse laplace transform of F(s)
$f\left(t\right)={L}^{-1}\left\{F\left(s\right)\right\}=\frac{14}{15}{e}^{3t}+\frac{2}{5}{e}^{-2t}-\frac{1}{3}$
$⇒{L}^{-1}\left\{F\left(s\right)\right\}=\frac{14}{15}{e}^{3t}+\frac{2}{5}{e}^{-2t}-\frac{1}{3}$