Laplace Transforms and third-order derivatives The question is to calculate the Laplace transform of $(1 + t.e^{-t})^3$. I know that this can be done using a property where the problem is of the form of $t.f(t)$. However, I seem to be messing up the third order derivative.  
This is a DIY exercise from the textbook and hence, no worked out solution.
I expanded the bracket $(1 + t^3e^{-3t} + 3te^{-t} + 3t^2.e^{-2t})$
and I mess up the third order derivative for $t^3.e^{-3t}$  
Can someone please how me how that is done ?

Hints:
$$(1+t e^{-t})^3 = 1 + 3 t e^{-t} + 3 t^2 e^{-2 t} + t^3 e^{-3 t}$$
$$\int_0^{\infty} dt \, t^k \, e^{-s t} = (-1)^k \frac{d^k}{ds^k} \frac{1}{s} = \frac{1}{s^{k+1}}$$
Now work out
$$\int_0^{\infty} dt \, e^{-s t} + 3 \int_0^{\infty} dt\,t \, e^{-(s+1) t}+ 3 \int_0^{\infty} dt\,t^2 \, e^{-(s+2) t}+  \int_0^{\infty} dt\,t^3 \, e^{-(s+3) t}$$
