So the question is find the inverse of $\dfrac{1}{s^2(s^2+\omega^2)}$. And here is the solution. I have no idea why its done this way. I would think to take a partial fraction of the form $\dfrac{a}{s}+\dfrac{b}{s^2}+\dfrac{c}{s^2+\omega^2}$ and then solve that, but that seems too complicated. This textbook solution looks like some brute force factorization specific to this problem only. Is there a general concept or way to tackle these kinds of problems thanks.
Let $F(s) = H(s) G(s)$ with $H(s) = \frac{1}{s^2}$ and $G(s) = \frac{1}{s^2 + \omega^2}$
By the convolution theorem $\mathcal{L}^{-1}\big(H(s) G(s)\big) = (h * g)(t) = \int_0^t \ h(\tau) g(t-\tau) \ d\tau$
From a table, $h(\tau) = \mathcal{L}^{-1}( 1/s^2 ) = \tau \ u(\tau)$ where $u(\tau)$ is the unit step function. $g(\tau) = \mathcal{L}^{-1}\big( 1/(\omega^2+ s^2 )\big) = \frac{\sin{\omega \tau}}{\omega} u(\tau)$.
$$(h * g)(t) = \int_0^t \ \tau \ u(\tau) \frac{\sin{\omega (t - \tau)}}{\omega} u(t-\tau) \ d\tau$$
Since $\tau$ is always positive $u(\tau) = 1$ on this domain. Similarly, since $t > \tau$ on this domain, $u(t-\tau) = 1$. The resulting integral can be done easily.
$$(h * g)(t) = \int_0^t \ \tau \ \frac{\sin{\omega (t - \tau)}}{\omega} \ d\tau = \frac{1}{\omega^2} (t - \frac{\sin{\omega t}}{\omega})$$