$\displaystyle L^{-1}\left[\frac{1}{s-1}\right]=\ \ e^t \\[3ex] \displaystyle L^{-1}\left[\frac{1}{s^2+4}\right]=\frac{1}{2}\sin{2t} \\[2ex] $

$\displaystyle \ By\ Convolution\ Theorem, \\[2ex] \displaystyle L^{-1}\left\{\frac{1}{(s-1)(s^2+4)}\right\}=\frac{1}{2}\left[\ \sin{2t}\times e^t\right] \\[2ex] \displaystyle \sin{2t}*e^t=\ \int_0^t\sin{2u}e^{t-u}du\ \\[2ex] \displaystyle =e^t\int_0^te^{-u}\ \sin{2u}\ du \\[2ex] \displaystyle =\frac{e^t}{\left(1^2+2^2\right)}{\left[-e^{-u}\sin{2u}-2e^{-u}\cos{2u}\right]}_0^t \\[2ex] \displaystyle =\frac{e^t}{5}\left[-e^{-t}\sin{2t}-2e^{-t}\cos{2t}-0+2\left(1\right)\left(1\right)\right] \\[2ex] \displaystyle =\frac{1}{5}[-\sin{2t}-2\cos{2t}+2e^t]\ \ \\[2ex] \displaystyle \therefore{},\ \ L^{-1}\left\{\frac{1}{(s-1)(s^2+4)}\right\}=\frac{1}{10}[-\sin{2t}-2\cos{2t}+2e^t]\ \ \ \ \\[2ex]$