Closed form of $\int_0^1\int_0^1\int_0^1\frac{\left(1-x^y\right)\left(1-x^z\right)\ln x}{(1-x)^3}\,\mathrm dx\;\mathrm dy\;\mathrm dz$ While trying to find several references to answer Pranav's problem, I encounter the following multiple integrals

$$I=\int_0^1\int_0^1\int_0^1\frac{\left(1-x^y\right)\left(1-x^z\right)\ln x}{(1-x)^3}\,\mathrm dx\;\mathrm dy\;\mathrm dz$$

Question :
Does the above integral have a closed form? If it has a closed form, how to obtain it?

According to Chriss'sis in the chat room, the integral has a nice closed form
$$\frac{2\ln 2 \pi -2 \gamma -5}{4}$$
but no proof given so far. Since the term $\ln 2 \pi$ appears in the close form, I have a hunch that it would involve Stirling's approximation. After trying to evaluate it for hours, I think it is about time to ask it on the main page. Here is my attempt so far:
\begin{align}
I&=\int_0^1\int_0^1\left(1-x^y\right)\;\mathrm dy\int_0^1\left(1-x^z\right)\;\mathrm dz\frac{\ln x}{(1-x)^3}\,\mathrm dx\\[7pt]
&=\int_0^1\frac{\ln x+1-x}{\ln x}\cdot\frac{\ln x+1-x}{\ln x}\cdot\frac{\ln x}{(1-x)^3}\,\mathrm dx\\[7pt]
&=\int_0^1\frac{\left(\ln x+1-x\right)^2}{(1-x)^3\ln x}\,\mathrm dx\\
\end{align}
I'm stuck in the latter expression. I'm thinking of the following parametric integral
$$I(s)=\int_0^1\frac{x^s\left(\ln x+1-x\right)^2}{(1-x)^3\ln x}\,\mathrm dx\quad\implies\quad I'(s)=\int_0^1\frac{x^s\left(\ln x+1-x\right)^2}{(1-x)^3}\,\mathrm dx$$
then expanding the quadratic term and using series representation of
$$\frac{1}{(1-x)^3}=\frac{1}{2}\sum_{n=1}^\infty n(n-1)x^{n-2}$$ 
but the calculation would be cumbersome. Is there a better way? Thanks in advance for your help.
 A: Continuing from Chris's sis answer, consider the partial sum:
$$\frac{1}{2}\sum_{n=1}^m n(n+1)\left(\ln(n)-2\ln(n+1)+\ln(n+2)\right)+1-\frac{1}{n}$$
$$=\frac{m-H_m}{2}+\frac{1}{2}\sum_{n=1}^m\left( \ln\left(\frac{n}{n+1}\right)^{n(n+1)}+\ln\left(\frac{n+2}{n+1}\right)^{n(n+1)}\right)$$
$$=\frac{m-H_m}{2}+\frac{1}{2}\left(\ln\left(\dfrac{\displaystyle \prod_{k=1}^m k^{2k}}{(m+1)^{m(m+1)}}\right)+\ln\left(\dfrac{(m+2)^{m(m+1)}}{\displaystyle \prod_{k=1}^m (k+1)^{2k}}\right)\right)$$
$$=\frac{m-H_m}{2}+\frac{1}{2}\ln\left(\frac{(m+2)^{m(m+1)}}{(m+1)^{m(m+1)}}\cdot \frac{(m!)^2}{(m+1)^{2m}}\right)$$
Use Stirling's approximation and rewrite the expression as:
$$\frac{1}{2}\left(\ln(2\pi)-H_m+\ln m+\ln\left(\left(\frac{m+2}{m+1}\right)^{m(m+1)}\left(\frac{m}{m+1}\right)^{2m}\frac{1}{e^m}\right)\right)$$
I still need to evaluate the logarithmic limit by hand but wolfram alpha gives $-5/2$ i.e the final result is:
$$\frac{1}{2}\left(\ln(2\pi)-\gamma-\frac{5}{2}\right)$$
A: Using series representation you specified above, I got that your integral gets reduced to
$$\sum _{n=1}^{\infty } \frac{(n+1) n^2 (\log (n)-2 \log (n+1)+\log (n+2))+n-1}{2 n}$$
Can you take it from here?
