Solution Verification - Integral of infinite power tower

Question

$$\int x^{x^{\cdot^{\cdot^\cdot}}}dx=\int -\frac{W(-\ln(x))}{\ln(x)}dx$$ Let $u=-\frac{W(-\ln(x))}{\ln(x)}$, $\frac{du}{dx}=\frac{u^2}{(1-\ln(u))u^{\frac{1}{u}}}\Rightarrow dx=\frac{(1-\ln(u))u^{\frac{1}{u}}}{u^2}du$. This derivative is not trivial but I don't have enough space to solve it here. I had a solution verification question were I solved this a while back Link for question. Many youtubers have also covered this. Using this information we can continue our integral: $$\int (1-\ln(u))u^{\frac{1}{u}-1}du$$ Let $v=\frac{1}{u}, \frac{dv}{du}=-v^2\Rightarrow du=-\frac{1}{u^2}$ $$-\int (1+\ln(v))v^{-v-1}=-\int \sum_{n=0}^{\infty}\frac{(1+\ln(v))(-1)^n(v+1)^n\ln(v)}{n!} dv=$$$$-\sum_{n=0}^{\infty}\sum_{k=0}^{n}\int\frac{(1+\ln(v))(-1)^nv^k\ln(v)^n}{k!(n-k)!}dv$$ Now we need to evaluate this integral:$$\int(1+\ln(v))(-1)^nv^k\ln(v)^ndv=(-1)^n\int v^k\ln(v)^{n+1}dv+(-1)^n\int v^k\ln(v)^ndv$$ Now we must solve two integrals:$$\int v^k\ln(v)^{n+1}dv$$Let $w=\ln(v)$, $dv=v dw$, $v^{k+1}=e^{(k+1)w}$, $ln(v)^{n+1}=w^{n+1}$ $$\int w^{n+1}e^{(k+1)w} dw$$ Let $t=w^{n+2}, dt= \frac{u^{-n-1}}{n+2}$ $$\frac{1}{n+2}\int e^{(k+1)t^{\frac{1}{n+2}}}dt$$ Let $a=(k+1)^{n+2}t, dt=(k+1)^{-n-2}da$ $$\frac{1}{(n+2)(k+1)^n(k^2+2k+1)}\int e^{a^{\frac{1}{n+2}}}=\frac{(-1)^{n+1}\Gamma(n+2, -a^{\frac{1}{n+2}})}{(k+1)^n(k^2+2k+1)}$$ $$=\dfrac{\left(-1\right)^{n+1}\Gamma(n+2,-\left(k+1)\ln\left(v\right)\right)}{(k+1)^n(k^2+2k+1)}$$ Now we have to evaluate one more integral: $$\int v^{n}\ln(v)^n dv$$ Let $b=\ln(v), dv= v db, v^{k+1}=e^{(k+1)b}, \ln(v)^{n}=b^n$ $$\int b^n e^{(k+1)b}db$$ Let $q=b^{n+1}, db=\frac{1}{(n+1)b^n}dq$ $$\frac{1}{n+1}\int e^{(k+1)q^{\frac{1}{n+1}}}$$ Let $m=(k+1)^{n+1}q, dq=(k+1)^{-n-1}dm$ $$\frac{1}{(n+1)(k+1)^{n+1}}\int e^{m^{\frac{1}{n+1}}}dm=(k+1)^{-n-1}(-1)^n\Gamma(n+1, -x^{\frac{1}{n+1}})=$$ $$(k+1)^{-n-1}(-1)^n\Gamma(n+1, -(k+1)\ln(v))$$ Now we have the solution to our integral: $$-(k+1)^{-n-2}(\Gamma(n+2, -(k+1)ln(v)))+(-k-1)\Gamma(n+1,-(k+1)ln(v))$$ Plugging this in to our original integral we get our solution: $$-\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{(k+1)^{-n-2}\Gamma(n+2, -(k+1)ln(v))+(-k-1)\Gamma(n+1,-(k+1)ln(v))}{k!(n-k)!}=$$ $$-\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{(k+1)^{-n-2}\Gamma(n+2, (k+1)W(-\ln(x)))+(-k-1)\Gamma(n+1, (k+1)W(-\ln(x)))}{k!(n-k)!}$$ Unfortunately I don't have any software to check if this is correct. Is it correct?