$\displaystyle Assuming\ the\ question\ to\ be\ \ \\[2ex] \displaystyle L^{-1}\left[\ {\tanh}^{-1}{\left(s\right)}\right]\ \ \\[2ex] \displaystyle Let,\ p(s)=\ \ {\tan h}^{-1}s \\[2ex] \displaystyle =\frac{1\ }{2}\log{\left(\frac{1+s}{1-s}\right)} \\[2ex] $

$\displaystyle =\frac{1}{2}[log\left(s+1\right)-log\left(1-s\right)] \\[2ex] \displaystyle \therefore{}2p'(s)=\ \frac{1}{s+1}-\frac{1}{1-s}\bullet{}-1=\frac{1}{s+1}-\frac{1}{s-1} \\[2ex] $

$\displaystyle Using\ Formula, \\[2ex] \displaystyle L^{-1}\left[F\left(s\right)\right]=\ -\frac{1}{t}\ L^{-1}[F^{'}\left(s\right)] \\[2ex] \displaystyle \therefore{}L^{-1}\left[\ {tanh}^{-1}s\ \ \right]=-\frac{1}{2t}L^{-1}[\frac{1}{s+1}-\frac{1}{s-1}] \\[2ex] \displaystyle =-\frac{1}{2t}(e^{-t}-e^t) \\[2ex] \displaystyle =\frac{1}{2t}(e^t-e^{-t}) \\[2ex] \displaystyle \therefore{},\ L^{-1}[{\tanh}^{-1}s]=\frac{1}{t}\times{}\sin ht\ \\[2ex] \displaystyle \frac{se^{-2s}}{(s^2+2s+2)} \\[2ex] $

$\displaystyle Consider,\\[2ex] \displaystyle \frac{s}{\left(s^2+2s+2\right)}=\frac{\left(s+1\right)-1}{{\left(s+1\right)}^2+1^2}\ \ \\[2ex] \displaystyle \because{},\ L\left[e^{at}f\left(t\right)\right]=F\left(s-a\right),\ \\[2ex] \displaystyle \therefore{},\ L^{-1}\left[F\left(s-a\right)\right]=e^{at}\ L^{-1}\left[F\left(s\right)\right] \\[2ex] \displaystyle \therefore{},L^{-1}\left[\frac{s}{\left(s^2+2s+2\right)}\ \right]=e^{-t}L^{-1}\left[\frac{s-1}{s^2+1}\right] \\[2ex] \displaystyle Consider,\\[2ex] \displaystyle L^{-1}\left[\frac{s-1}{s^2+1}\right]=\ L^{-1}\left[\frac{s}{s^2+1}\right]-L^{-1}\left[\frac{1}{s^2+1}\right]=\cos{t\ }-\sin t \\[2ex] \displaystyle \therefore{},\ \ L^{-1}\left[\frac{s}{\left(s^2+2s+2\right)}\ \right]=e^{-t}\left(\cos t-\sin t\right) \\[2ex] $

$\displaystyle now,\\[2ex] L\left[f\left(t-a\right)u\left(t-a\right)\right]=e^{-as}F\left(s\right),\ \\[2ex] \displaystyle \therefore{},\ L^{-1}\left[e^{-as}F\left(s\right)\right]=\ f\left(t-a\right)u\left(t-a\right) \\[2ex] \displaystyle \therefore{},\ L^{-1}\left[\frac{se^{-2s}}{\left(s^2+2s+2\right)}\ \right]=e^{-(t-2)}\left(\cos{(t-2)}-\sin{(t-2)}\right)\ u(t-2) \\[2ex] $