Let,

$\phi_{1}(s) = \frac{1}{s - 1}$

$\phi_{2}(s) = \frac{1}{s^2 + 4}$

Convolution Theorem:

If $ L^{-1}[f(s)] = f(t)$ & $L^{-1}[g(s)]*g(t)$ then

$L^{-1}[f(s)g(s)] = \int\limits_{0}^{t} f(u) g(t - u) du$

$\therefore$ Taking Inverse Laplace Transform of $\phi_{1}(s)$ and $\phi_{2}(s)$ we get,

$L^{-1}[\phi_{1}(s)] = L^{-1}\bigg[\frac{1}{s - 1}\bigg] = e^t$

$L^{-1}[\phi_{2}(s)] = L^{-1}\bigg[\frac{1}{s^2 + 4}\bigg] = L^{-1}\bigg[\frac{1}{2}sin2t\bigg]$

$\therefore L^{-1}[\phi_{2}(s)\phi_{1}(s)] = \frac{1}{2}\int\limits_{0}^{t}sin2ue^{t - u}du$

$\therefore L^{-1}[\phi_{2}(s)\phi_{1}(s)] = \frac{e^t}{2}\int\limits_{0}^{t}sin2ue^{-u}du$

$\therefore L^{-1}[\phi_{2}(s)\phi_{1}(s)] = \frac{e^t}{2}\bigg[\frac{e^{-u}}{(-1)^2 + 2^2}(-1sin2u - 2cos2u)\bigg]_{0}^{t}$

$\therefore L^{-1}[\phi_{2}(s)\phi_{1}(s)] = \frac{e^t}{2}\bigg[\frac{e^{-t}}{5}(-sin2t-2cos2t)\bigg]_{0}^{t}$

$\therefore L^{-1}[\phi_{2}(s)\phi_{1}(s)] = \frac{e^t}{2}\bigg[\frac{e^{-t}}{5}(-sin2t-2cos2t) - \frac{1}{5}(-2)\bigg]$

$\therefore L^{-1}[\phi_{2}(s)\phi_{1}(s)] = \frac{-1}{10}(sin2t + 2cos2t)\frac{2}{5}e^t$

$\therefore L^{-1}[\phi_{2}(s)\phi_{1}(s)] = \frac{1}{10}[se^t - (sin2t + 2cos2t)]$

$\therefore L^{-1}\bigg[\frac{1}{(s - 1)(s^2 + 4)}\bigg] = \frac{1}{10}[2e^t - (sin2t + 2cos2t)]$