# Take the Inverse Laplace Transform of the function.F(s)=\frac{1}{s(s^2+2s+2)}

Take the Inverse Laplace Transform of the function.
$F\left(s\right)=\frac{1}{s\left({s}^{2}+2s+2\right)}$
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Khribechy

$F\left(s\right)=\frac{1}{s\left({s}^{2}+2s+2\right)}=\frac{1}{s\left({\left(s+1\right)}^{2}+{\left(1\right)}^{2}\right)}=\frac{1}{s}\frac{1}{\left({\left(s+1\right)}^{2}+{\left(1\right)}^{2}\right)}$
$⇒F\left(s\right)=L\left\{1\right\}L\left\{{e}^{-t}\mathrm{sin}\left(t\right)\right\}$
$⇒{L}^{-1}\left\{F\left(s\right)\right\}=f\left(t\right)=1\cdot \left({e}^{-t}\mathrm{sin}t\right)$
$⇒f\left(t\right)={\int }_{0}^{t}{e}^{-u}\mathrm{sin}\left(u\right)1du={\left(\frac{{e}^{-k}}{2}\left[-1\mathrm{sin}\left(u\right)-1\mathrm{cos}\left(u\right)\right]\right)}_{0}^{t}$
$={\left(-\frac{{e}^{-4}}{2}\left[\mathrm{sin}\left(u\right)+\mathrm{cos}\left(u\right)\right]\right)}_{0}^{t}$
$=-\frac{1}{2}\left[{e}^{-t}\left(\mathrm{sin}\left(t\right)+\mathrm{cos}\left(t\right)\right)-1\right]$
$⇒f\left(t\right)=\frac{1}{2}\left[1-{e}^{-t}\left(\mathrm{sin}\left(t\right)+\mathrm{cos}\left(t\right)\right)\right]$