I got this from working on the integral of
$$x^x dx$$
and with
$$u=x^x \Rightarrow du=x^x\cdot \ln(ex)$$
I got the integral of
$$\frac{u}{(x^x\cdot \ln(ex))}du$$
which becomes
$$\frac{1}{\ln(ex)} du$$
which using the fact that
$$x=e^{W(\ln(u))}$$
(when x is real and greater than one)
I got $$\frac{1}{\ln(e^{W(\ln(u)+1})} du$$ which simplifies to $$\frac{1}{W(\ln(u))+1} du$$ which using $$v=\ln(u) \Rightarrow dv=\frac{1}{u} du$$ or $$dv\cdot u=du$$ or $$dv\cdot e^v=du$$ so then I got $$\frac{1}{W(v)+1}\cdot e^v dv$$ so that is how I got the integral of $$\frac{e^v}{W(v)+1}dv $$
note: $W(z)$ is the Lambert W function of z
http://mathworld.wolfram.com/LambertW-Function.html
If you find a mistake in one of my steps please put it in the comments and I will fix it as soon as I can
So, my final question is: What is the integral of $$\frac{e^v}{W(v)+1}dv $$
Note: in case it’s useful, according to Wolfram Alpha the integral of $\frac{1}{W(v)+1}$ is $\frac{v}{W(v)}$