$\displaystyle i)\ Let\ \varnothing{}\left(s\right)\\[2ex] \displaystyle =\log{\left(\frac{s-2}{s-3}\right)} \\[3ex] \displaystyle =\log{\left(s-2\right)}-\log{\left(s-3\right)} \\[2ex] \displaystyle \therefore{}{\varnothing{}}^{'}\left(s\right)=\frac{1}{s-2}-\frac{1}{s-3} \\[2ex] $

$\displaystyle Using\ Formula, \\[2ex] \displaystyle \\[2ex] \displaystyle L^{-1}\left[\varnothing{}\left(s\right)\right]=-\frac{1}{t}L^{-1}\left[{\varnothing{}}^{'}\left(s\right)\right] \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}L^{-1}\left[\log{\left(\frac{s-2}{s-3}\right)}\right]\\[3ex]=-\frac{1}{t}L^{-1}\left[\frac{1}{s-2}-\frac{1}{s-3}\right]\ \\[2ex] \displaystyle =-\frac{1}{t}\left(e^{2t}-e^{3t}\right) \\[2ex] \displaystyle \\[2ex] \displaystyle =\frac{1}{t}\left(e^{3t}-e^{2t}\right) \\[2ex] $

$ii)\ L^{-1}\left[\left(\dfrac{s+1}{s^2-4}\right)\right]\\[3ex] \displaystyle =L^{-1}\left[\left(\frac{s}{s^2-2^2}\right)\right]+L^{-1}\left[\left(\frac{1}{s^2-2^2}\right)\right] \\[3ex] =\cosh{2t+\frac{1}{2}\sinh{2t}} $