Find inverse Laplace transform.

$ L^{-1}\left [ \log\left ( \dfrac{s^2+1}{s(s+1)} \right ) \right ] $

$L^{-1}\ log\Bigg[\dfrac{s^2+1}{s(s+1)}\Bigg]=-\dfrac{1}{t}L^{-1}\Bigg[\dfrac{d}{ds}\Bigg(log\dfrac{s^2+1}{s(s+1)}\Bigg)\Bigg]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\dfrac{1}{t}L^{-1}\Bigg[\dfrac{d}{ds}\Bigg(log(s^2+1)-logs-log(s+1)\Bigg)\Bigg]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\dfrac{1}{t}L^{-1}\Bigg[\dfrac{2s}{s^2+1}-\dfrac{1}{s}-\dfrac{1}{s+1}\Bigg]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\dfrac{1}{t}\Bigg[2L^{-1}\dfrac{s}{s^2+1}-L^{-1}\Bigg( \dfrac{1}{s}\Bigg)-L^{-1}\dfrac{1}{s+1}\Bigg]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\dfrac{1}{t}\ [2\ cos\ t-1-e^{-t}]$