i)Let $\varphi(s) = \log \left [ \dfrac {s^2 + a^2}{\sqrt{s+b}} \right ]$

$= \log (s^2 +a^2) - \log (s+b)^{1/2}$

$= \log (s^2 + a^2)-\dfrac {1}{2} \log (s+b)$

$\therefore ;\varphi'(s) = \dfrac {1}{s^2 + a^2} \cdot 2s- \dfrac {1}{2}\cdot \dfrac {1}{s+b}$

Using formula,

$L^{-1} \big [ \varphi (s) \big] = - \dfrac {1}{t} L^{-1} \Big [ \varphi' (s) \Big]$

$L^{-1} \left \{ \log \left[ \dfrac {s^2 + a^2}{\sqrt{s+b}} \right ] \right \} = -\dfrac {1}{t} L^{-1} \left [ \dfrac {2s}{s^2 + a^2} - \dfrac {1}{2} \cdot \dfrac {1}{s+b} \right ]$

$= \dfrac {1}{t} \left \{ 2L^{-1} \left [ \dfrac {s}{s^2 +a^2} \right ] - \dfrac {1}{2}L^{-1} \left [ \dfrac {1}{s+b} \right ] \right \}$

$= -\dfrac {1}{t} \left \{ 2 \cos at - \dfrac {1}{2} e^{-bt} \right \}$

$= \dfrac {1}{2t} \big \{e^{-bt} - 4 \cos at \big\}$

ii)Let $\varphi_1 (s) = \dfrac {1}{s^3}; \ \ \varphi_2(s) = \dfrac {1}{s-1}$

$\therefore f_1 (t) = L^{-1} \left [ \dfrac {1}{s^3} \right ] = \dfrac {t^2}{2}$

And, $f_2 (t) = L^{-1} \left [ \dfrac {1}{s-1} \right ]= e^t$

By Convolution theorem,

$\displaystyle L^{-1} \Big [ \varphi_1 (s)\varphi_2 (s) \Big] = \int^{t}_0 f_1 (u)f_2 (t-u)du $

$\therefore \displaystyle L^{-1} \left [ \dfrac {1} {s^3} \times \dfrac {1}{s-1} \right ] = \int^{t}_0 \dfrac {u^2}{2} \times e^{(t-u)}du $

$\displaystyle =\dfrac {1}{2} \int^t_0 u^2 \ e^t \ e^{-u} du $

$\displaystyle = \dfrac {1}{2}e^{t} \int^t_0 u^2 \ e^{-u} du$

$= \dfrac {1}{2} e^t \left [ u^2 \times \dfrac {e^{-u}}{-1} -2u \times \dfrac {e^{-u}}{(-1)^2} + 2\times \dfrac {e^{-u}}{(-1)^3} \right ]^t_0$

$= \dfrac {1}{2}e^{t} \Big\{ \Big [ -t^2 e^{-t} - 2te^{-t} - 2e^{-t} \Big] \Big\} - \Big[ 0-0-2\times 1 \Big ]$

$= \dfrac {1}{2} e^{t} \Big \{ -e^{-t} \Big [t^2 + 2t +2 \Big ] +2 \Big\}$

$= \dfrac {1}{2} \Big\{ - \Big [t^2 + 2t +2 \Big ]+2e^t \Big \}$

$= \dfrac {-1}{2} \Big [ t^2 +2t + 2\Big ] + e^t$

$\therefore L^{-1} \left [ \dfrac {1}{s^3 (s-1)} \right ] = e^t - \dfrac {1}{2}t^2 - t-1$