Solution.

Part 1

\(\displaystyle {H(s) = \frac {1}{(s^2 + 1)^2} = \frac {1}{(s^2 + 1)}\cdot \frac {1}{(s^2 + 1)} = F(s)\cdot G(s)}\) where \(\displaystyle {F(s) = \frac {1}{s^2 + 1} = G(s)}\) and so \(f(t) = \sin t = g(t)\).

\begin {align*} \mathcal {L}^{-1}\big [H(s)\big ] & = \mathcal {L}^{-1}\big [F(s)\hspace {0.1cm} G(s)\big ]\\ & = f(t)*g(t)\\ & = \int ^t_0 f(\tau ) g(t-\tau ) d\tau \end {align*}

\begin {align*} \text {Hence}\hspace {0.5cm} \mathcal {L}^{-1}\big [H(s)\big ] & = \int ^t_0\sin (\tau ) \sin (t-\tau ) d\tau \\\\ & = \frac {-1}{2}\int ^t_0 \big [\cos t - \cos (2\tau -t)\big ]d\tau \\\\ & = \frac {-1}{2} \cos t\big [\tau \big |^t_0 +\frac {1}{4} \sin \big (2\tau -t\big )\big |^t_0\\\\ & = \frac {-t}{2}\cos t+ \frac {1}{4}\big [\sin t- \sin (-t)\big ]\\ & = \frac {-t}{2}\cos t + \frac {2}{4}\sin t\\ & = \frac {1}{2}\big (\sin t - t\cos t\big ) \end {align*}

\[\therefore \hspace {0.5cm} \mathcal {L}^{-1}\Bigg [\dfrac {1}{(s^2 + 1)^2}\Bigg ] = \frac {1}{2}\big (\sin t - t\cos t\big )\]

Part 2

\(\displaystyle {H(s) = \frac {1}{s^2 (s-a)} = \frac {1}{s^2}\cdot \frac {1}{s-a}= F(s)\cdot G(s)}\) where \(F(s)= \dfrac {1}{s^2}\) and \(G(s) = \dfrac {1}{s -a}\)

\(\implies \hspace {0.5cm}f(t) = t\) and \(g(t) = e^{at}\) \begin {align*} \mathcal {L}^{-1}\big [H(s)\big ] & = \mathcal {L}^{-1}\big [F(s)\hspace {0.1cm} G(s)\big ]\\ & = \int ^t_0 \tau e^{a(t-\tau )} d\tau \\ & = e^{at} \int ^t_0\tau e^{-a\tau }d \tau \\ & = \frac {1}{a^2}\big (e^{at} -at -1\big )\\\\ \end {align*}