How to find the Laplace Transform of $t^2sin(t)$ Using the rule: $$\mathcal{L}(t^nf(t))=(-1)^n\frac{d^n}{ds^n}F(s)$$                       
where in this case
$$f(t)=\sin(t),\,\,\,\,\,\,\,\,\,\mathcal{L}(\sin(t))=F(s)=\frac1{s^2+1},\,\,\,\,\,\,\,\,\,\,n=2.$$
Find the 2nd derivative of F(s):
$$\frac{d^2}{ds^2}\Big(\frac1{s^2+1}\Big)=\frac{6s^2-2}{(s^2+1)^3}$$
The transform:
$$\mathcal{L}(t^2sin(t))=(-1)^2\frac{6s^2-2}{(s^2+1)^3}$$
$$= \frac{6s^2-2}{(s^2+1)^3}$$

$L\{tf(t)\} = \int_0^{\infty} tf(t) e^{-st} dt$
if we integrate both sides with respect to $s.$
$\int L\{tf(t)\} ds = -\int_0^{\infty} f(t) e^{-st} dt = -L\{f(t)\}$
And then we can differentiate both sides (with respect to $s$).
$L\{tf(t)\} = -\frac {d}{ds} L\{f(t)\}$
$L\{t^2\sin t\}=\frac{d^2}{ds^2} \frac {1}{s^2+1}$
