I seem to have gotten stuck on this problem:
Ive tried partial integration only to end up with $\frac{2}{s+4} \cdot t^2 \int_0^\infty \frac{2}{s+4} 2t\cdot dt$ which would be $-\infty$ on the Real axis i believe.
Shouldn't be right though. Mathematica is giving $\frac{2s(-12+s^2)}{(4+s^2)^3}$ as a solution.
You can also use this result: $$\mathcal {L}(t^nf(t))=(-1)^n\dfrac{d^n}{ds^n}(\mathcal{L}(f(t))(s))$$ In your case it means: $$\mathcal {L}(t^2\cos (2t) )=\dfrac{d^2}{ds^2}(\mathcal{L}(\cos (2t) )(s))$$ It's easy to find the Laplace transform of $cos$ function. And differentiate twice. $$\mathcal {L}(t^2\cos (2t) )=\dfrac{d^2}{ds^2}\left(\dfrac s {s^2+4}\right)$$ Differentiate twice w.r.t to the variable $s$ the RHS. You will find Mathematica's answer. $$\mathcal {L}(t^2\cos (2t) )=\dfrac{d}{ds}\dfrac {4-s^2} {(s^2+4)^2}$$ $$\mathcal {L}(t^2\cos (2t) )=2s\dfrac {(s^2-12)} {(s^2+4)^3}$$