Sum of $\sum_{n \geq 1} \frac{(\ln x +1)^n}{n^n}$
Solution 1:
Only a hint:
For $\,a\geq 0\,$ it's $$\sum\limits_{k=0}^\infty\frac{x^k}{(ak+1)^{k+1}}=\int\limits_0^1 t^{-xt^a}dt $$
and the base for this result is calculating $\enspace\int\limits_0^1 t^b(\ln t)^cdt\enspace$.
Therefore with $a:=1$ and $x$ substituted by $\ln x+1$ we get $$\sum_{n \geq 1} \frac{(\ln x +1)^n}{n^n}=(\ln x+1)\int\limits_0^1 t^{-t(\ln x +1)}dt \enspace.$$