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.

$$
\int_0^\infty \left(\frac {2x^3}{3K}\right) e^{-x} \, dx = \frac 2 {3K} \int_0^\infty x^3 e^{-x} \, dx. 
$$
The integral here is $\displaystyle\int_0^\infty x^3 e^{-x}\, dx;$ the fraction $\dfrac 2 {3K}$ can be attended to after evaluating the integral.
You say you did integration by parts multiple times. Now suppose you had had instead $$\int_0^\infty x^{50} e^{-x} \, dx.$$ Then you might think you'd have to do integration by parts a much larger number of times. However, all of those iterations of integration by parts are the same, thus:
\begin{align}
\int_0^\infty x^{50} \Big( e^{-x} \, dx\Big) & = \int u\, dv = uv - \int v\, du \\[10pt]
& = \left. x^{50} (-e^{-x}) \vphantom{\frac \sum1}\, \right|_0^\infty - \int_0^\infty (-e^{-x}) \Big( 50 x^{49}\, dx\Big) \\[10pt]
& = 50\int_0^\infty x^{49} e^{-x}\, dx.
\end{align}
You're back to the same integral, except that the exponent has been reduced from $50$ to $49,$ and you also have a factor of $50$ in front of the integral. So doing it again reduces the exponent from $49$ to $48$ and you get a factor of $49$ in front of the integral:
$$
50 \cdot 49 \int_0^\infty x^{48} e^{-x}\,dx.
$$
Then do it again:
$$
50\cdot49\cdot48\int_0^\infty x^{47} e^{-x}\,dx.
$$
And so on:
$$
50\cdot49\cdot48\cdot47\cdot46\cdots\cdots3\cdot2\cdot1\int_0^\infty x^0 e^{-x}\, dx
$$
and you don't need integration by parts for that last one.
So once you see the pattern, you've got it.
