Trick to calculate integrals such as $\int_0^\infty \left(\frac {2x^3}{3K}\right) e^{-x}dx$? Besides integration by parts, is there any trick to calculate integrals such a$$\int_0^\infty \left(\frac {2x^3}{3K}\right) e^{-x}dx=1 $$
I got $K=4$ by using integration by parts multiple times, but I think there should be some sort of formula to solve this kind of integrals.

HINT: the indefinite integral is given by $$-2/3\,{\frac { \left( {x}^{3}+3\,{x}^{2}+6\,x+6 \right) {{\rm e}^{-x}}
}{K}}
+C$$
