Evaluate ∫01(1ln⁡x+11−x)2dx

A bit late to the party, but here is a slightly different approach that doesn't use any series, Stirling's approximations or anything, though does make use of some values of the zeta function.
First substitute ${\displaystyle x=e^{-y}}$ to get
${\displaystyle {\int }_0^{\mathrm{\infty }}\left(\frac{1}{1-e^{-y}}\right)-\frac{1}{y}{)}^2{e}^{-y}dy}$
The integrand is awkward because when you multiply it out, the pieces separately diverge. Multiplying it by ${\displaystyle y^z}$ regulates the singularity at 0 but also gives standard integrals. So for ${\displaystyle Re\left(z\right)>1}$ we have (by parts)
${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{1}{{(1-e^{-y})}^2}{e}^{-y}{y}^{z}dy=z{\int }_0^{\mathrm{\infty }}\frac{e^{-y}}{1-e^{-y}}{y}^{z-1}dy}$
${\displaystyle =z\zeta \left(z\right)\mathrm{\Gamma }\left(z\right)}$
and
${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{1}{y(1-e^{-y}\}{e}^{-y}{y}^{z}dy=\zeta \left(z\right)\mathrm{\Gamma }\left(z\right)}}$
${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{1}{y^2}{e}^{-y}{y}^{z}dy=\mathrm{\Gamma }(z-1)}$
Putting the bits together:
${\displaystyle {\int }_0^{\mathrm{\infty }}{(\frac{1}{1-e^{-y}}-\frac{1}{y})}^2{e}^{-y}{y}^{z}dy}$
${\displaystyle =\mathrm{\Gamma }\left(z\right)(\frac{1}{z-1}+(z-2)\zeta \left(z\right))}$
The derivation was valid for ${\displaystyle Re\left(z\right)>1}$, but both sides of the above equation are analytic for ${\displaystyle Re\left(z\right)\succ 1}$ (RHS has a removable singularity at ${\displaystyle z=0}$), so by analytic continuation we may take the limit ${\displaystyle z\to 0}$. Near ${\displaystyle z=0,\mathrm{\Gamma }\left(z\right)=\frac{1}{z}+O\left(1\right)}$, and the final answer is the derivative of ${\displaystyle \frac{1}{z-1}+(z-2)\zeta \left(z\right)}$ at o, i.e.,
${\displaystyle -1+\zeta \left(0\right)-2{\zeta }^{\prime }\left(0\right)=\log \left(2\pi \right)-\frac{3}{2}}$

OK, I'm going to lay this out up to a sum, which will likely evaluate into whatever answer was provided above. This integral is subject to the same sorts of tricks that I did for another integral involving a factor of $1/\log x$ in the integral. The first piece is to let $x=e^{-y}$; the integral becomes
${\int }_0^{\mathrm{\infty }}dy{e}^{-y}(\frac{(e^{-y}-(1-y){)}^2}{y^2(1-e^{-y}{)}^2})$
Now Taylor expand the factor $(1-e^{-y}{)}^{-2}$, and if we can reverse the order of summation and integration, we get:
$\sum _{k=1}^{\mathrm{\infty }}k{\int }_0^{\mathrm{\infty }}dy(\frac{e^{-y}-(1-y){)}^2}{y^2})e^{-ky}$
The integral inside the sum is a bit difficult, although it is convergent. The way I see through it is to replace k with a continuous parameter $\alpha$ and differentiate with respect to $\alpha$ inside the integral twice (to clear the pesky $y^2$ in the denominator) to get a function
$$\begin{gathered} I(\alpha )={\int }_0^{\mathrm{\infty }}dy(\frac{(e^{-y}-(1-y){)}^2}{y^2})e^{-\alpha y} \\ \frac{d^{2}I}{d{\alpha }^2}={\int }_0^{\mathrm{\infty }}dy(e^{-y}-(1-y){)}^2{e}^{-\alpha y} \\ =\frac{1}{\alpha +2}-\frac{2}{\alpha +1}+\frac{2}{(\alpha +1{)}^2}+\frac{1}{\alpha }-\frac{2}{{\alpha }^2}+\frac{2}{{\alpha }^3} \end{gathered}$$
You integrate this twice to recover $\alpha$; the constants of integration may be shown to vanish by considering the limit as $a\to \mathrm{\infty }$. The original integral is then
$\sum _{k=1}^{\mathrm{\infty }}kI(k)$
where
$I(k)=(k+2)\log [\frac{k(k+2)}{(k+1{)}^2}]+\frac{1}{k}$
so the integral takes on the value
$\sum _{k=1}^{\mathrm{\infty }}[1+[(k+1{)}^2-1]\log (1-\frac{1}{(k+1{)}^2})]=\sum _{k=1}^{\mathrm{\infty }}[1+(k+1{)}^2\log (1-\frac{1}{(k+1{)}^2})]+\log 2$
The sum may be simplified by Taylor expanding the $\log$ term; note that the unit value cancels and we get that the integral equals
$\log 2+\sum _{k=2}^{\mathrm{\infty }}[1-k^2\sum _{m=1}^{\mathrm{\infty }}\frac{1}{m}(\frac{1}{k^2}{)}^m]=\log 2-\sum _{m=1}^{\mathrm{\infty }}\frac{1}{m+1}[\zeta (2m)-1]$
I have not yet evaluated this sum yet, but unless someone else does it before me, I will figure it out and come back.