How to solve the transforms below step by step
$$\mathscr{L}^{-1} \frac{1}{(s+ \lambda)^2- \omega^2} $$
$$\mathscr{L}^{-1} \frac{a(s+2 \lambda)+b}{(s+ \lambda)^2- \omega^2} $$
I found some transformed tables, but I couldn't understand where the result comes from, or how to reproduce the accounts step by step.
These come from the Laplace transforms of $e^{-\lambda t}\cosh(\omega t)$ and $e^{-\lambda t}\sinh(\omega t)$: $$ \mathcal{L}\left[e^{-\lambda t}\sinh(\omega t)\right] = \frac{1}{2}\left(\mathcal{L}\left[e^{-(\lambda - \omega) t}\right] - \mathcal{L}\left[e^{-(\lambda+\omega) t}\right]\right) = \frac{1}{2}\left(\frac{1}{s+\lambda-\omega} - \frac{1}{s+\lambda + \omega}\right) = \frac{\omega}{(s+\lambda)^2 -\omega^2} $$ $$ \mathcal{L}\left[e^{-\lambda t}\cosh(\omega t)\right] = \frac{1}{2}\left(\mathcal{L}\left[e^{-(\lambda - \omega) t}\right] + \mathcal{L}\left[e^{-(\lambda+\omega) t}\right]\right) = \frac{1}{2}\left(\frac{1}{s+\lambda-\omega} + \frac{1}{s+\lambda + \omega}\right) = \frac{s+\lambda}{(s+\lambda)^2 -\omega^2} $$ To solve those Laplace transforms, you simply need to write them as linear combinations of the above two transforms.