As it says in the title, I'd like to know how to solve the definite integral $\int_0^1 \log{\Gamma(x+1)}\mathrm dx$. Mathematica gives the answer $\frac{1}{2}\log (2\pi)-1$ but I have no idea how one would obtain that (also I feel like Stirling's Approximation is somehow involved since it seems to like the number $\log (2\pi)$ and concerns itself with the relationship between exponentials and factorials).
Motivation: This gives a pretty cool relation, even if it probably can't be used for anything. $$ \frac{\sqrt{2\pi}}{e}=e^{\int_o^1 \log(\Gamma(x+1))\mathrm dx}=\lim_{n\to\infty} e^{\frac{1}{n}\sum_{k=0}^n \log(\Gamma(\frac{k}{n}+1))}=\lim_{n\to\infty} \left(\prod_{k=0}^n\Gamma\left(\frac{k}{n}+1\right)\right)^{1/n} $$
TL;DR: How can we solve the integral, is the value related to Stirling, and if so, how?