How to prove that : $$ \int x^{n-1}W(x)dx = \frac {x^ne^{[-nW(x)]}[-nW(-x)]^{-n}[n\Gamma(n+1, -nW(x)- \Gamma(n+2, -nW(x))]} {n^2} $$ Where $W(x)$ is the Lambert-W function https://en.wikipedia.org/wiki/Lambert_W_function and $\Gamma$ is the incomplete gamma function.
I found this on wolfram alpha https://www.wolframalpha.com/input/?i=int+(x)%5E(n-1)W(x), but have no idea how it got this.
For those who are wondering why I want to know the proof of this integral it is because with this integral I can find the integral of the infinite tetration of x as :
$ \int x^{x^{x^{.^{.......}}}} dx = \sum_{n=0}^{\infty}-\frac {(-1)^n( \ln x)^ne^{[-nW(-\ln x)]}[-nW(-\ln x)]^{-n}[n\Gamma(n+1, -nW(-\ln x)- \Gamma(n+2, -nW(-\ln x))]} {(n!)n^2} $
Lol the answer is so long that it doesnt fit on one line.
You probably want to make the substitution $ue^u = x$,
$dx = (u+1)e^u du$ that would give you $\int u^{n-1} e^{(n-1)u} W(ue^u) (u+1)e^u du$.
We know that $W(ue^u) = u$
Hence our integral is
$\int u^{n} e^{nu}(u+1) du$.
What next?
Well $\Gamma(x) = \int t^{x-1} e^{-t}dt$ So our integral looks kind of like a gamma integral but not quite.
We will separate it into
$\int u^{n+1} e^{nu} du (1) + \int u^{n} e^{nu} du (2)$.
$t = -nu$
$(1) = \int (-1/n)^{n+1} t^{n+1} e^{-t} dt/(-n) = (-1/n)^{n+2} \Gamma(n+2)$
$(2) = \int (-1/n)^{n} t^{n} e^{-t} dt/(-n) = (-1/n)^{n+1} \Gamma(n+1)$
Final answer $(-1/n)^{n+2} \Gamma(n+2)+(-1/n)^{n+1} \Gamma(n+1)$
(tbc)