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:
$$\mathcal{L}(t^2sin(t))=(-1)^2\frac{6s^2-2}{(s^2+1)^3}$$
$$= \frac{6s^2-2}{(s^2+1)^3}$$

Hint:
$$\mathcal{L}(f(t))=\int_{0}^{\infty} f(t) e^{-st} \ dt.$$
Differentiate under the integral sign twice with respect to $s$ , and we get
$$\frac{\partial^2 \mathcal{L}(f(t))}{\partial s^2}=\int_{0}^{\infty} t^2 f(t) e^{-st} \ dt=\mathcal{L}(t^2 f(t)).$$ 
What happens when $f(t)=\sin(t)$ ? 
