Question

inverse laplace transform
\[\scr {L}^{-1}\left\{ln\left(\frac{s^2+2s+5}{s^2-4s+13}\right)\right\}\]

Answer 1

look at numbers 8 & 10

http://eqworld.ipmnet.ru/en/auxiliary/inttrans/LapInv7.pdf

Answer 2

Recalling that
\[\mathcal{L}^{-1}\left\{\frac{\mathrm{d}^n}{\mathrm{d}s^n}F(s)\right\}=(-t)^nf(t)\]
where \(F(s)=\mathcal{L}\{f(t)\}\), you have
\[\begin{align*}F(s)&=\ln\frac{s^2+2s+5}{s^2-4s+13}\\[2ex]
\frac{\mathrm{d}F}{\mathrm{d}s}&=\frac{46+16s-6s^2}{65+6s+10s^2-2s^3+s^4}\\[2ex]
&=-2\frac{3s^2-8s-23}{(s^2-4s+13)(s^2+2s+5)}\\[2ex]
&=2\left(\frac{s+1}{s^2+2s+5}-\frac{s-2}{s^2-4s+13}\right)\\[2ex]
&=2\left(\frac{s+1}{(s+1)^2+4}-\frac{s-2}{(s-2)^2+9}\right)
\end{align*}\]
The rest should be easy. Applying that formula from above, this gives you
\[\mathcal{L}^{-1}\left\{\ln\frac{s^2+2s+5}{s^2-4s+13}\right\}=\mathcal{L}^{-1}\left\{\frac{\mathrm{d}F}{\mathrm{d}s}\right\}=-tf(t)\]