Methods to calculate an approximation of $\int_0^1 \log\left(1+x^{\Gamma(x)}\right)\,dx,$ where $\Gamma(x)$ is the Gamma function While I was playing with Wolfram Alpha online calculator I wondered how calculate an approximation of $$I=\int_0^1 \log\left(1+x^{\Gamma(x)}\right)\,dx,\tag{1}$$ where $\Gamma(x)$ is the Gamma function. I am not interested specially to get a very good approximation, because Wolfram Alpha provide me one with code
int log(1+x^Gamma(x))dx, from x=0 to 1
I would like to know methods about how calculate an approximation of $I$. I suspect that to get a good approximation is required to do analysis about the Gamma function.

Question. How calculate using analysis an approximation of $$\int_0^1 \log\left(1+x^{\Gamma(x)}\right)\,dx?$$ Many thanks.

 A: Only a note.
You can use Riemann sums (e.g. https://en.wikipedia.org/wiki/Riemann_sum) for a first approximation.
Be $\enspace\displaystyle f(n):=\sqrt[n]{\prod\limits_{k=1}^n(1+(\frac{k}{n})^{\Gamma(\frac{k}{n})})} \,$ . Then we get $\enspace\displaystyle \int\limits_0^1 \ln(1+x^{\Gamma(x)})dx=\ln\lim\limits_{n\to\infty}f(n)\,$ . 
It's $\enspace\displaystyle \int\limits_0^1 \ln(1+x^{\Gamma(x)})dx =0,2831535412286936...$ 
and e.g. $\enspace\ln f(10000)\approx 0.283188199004388115\,$ .
Disadvantage: The convergence is very slow. 
A: First we note that within $(0,1)$ we have  $x^{\Gamma(x)}<x$.
From this we get immediately an upper bound for $I$:
$$\int_0^1 \log\left(1+x\right)\,dx=2\ln2-1=0.3862...$$ 
If we are not satisfied with this robust estimation then within $(0,1)$ a good approximation for $\Gamma(x)$ is $\frac {1}{x}$ 
So we get a better estimation for $I$:
$$\int_0^1 \log\left(1+x^{\frac {1}{x}}\right)\,dx$$ 
Since $0<x^{\frac {1}{x}}<1$ we can use Taylor series for the logarithm 
$$\int_0^1 \log\left(1+x^{\frac {1}{x}}\right)\,dx=f(1)-\frac {f(2)}{2}+...+\frac {(-1)^{(k+1)}f(k)}{k}+...$$  
where 
$$f(k)=\int_0^1 x^{\frac {k}{x}}d\,x$$  
For example $\frac {f(10)}{10}=0.0079...$ but since we calculate only an estimation i think it is needed fewer terms to get a satisfactory estimate.
