Integral $\int_0^1\frac{\ln x}{\left(1+x\right)\left(1+x^{-\left(2+\sqrt3\right)}\right)}dx$ There is a curious known integral:
$$\int_0^1\frac{\ln\left(1+x^{2+\sqrt{3\vphantom{\large3}}}\right)}{1+x}dx=\frac{\pi^2}{12}\left(1-\sqrt{3\vphantom{\large3}}\right)+\ln \left(1+\sqrt{3\vphantom{\large3}}\right)\ln2.$$
If we consider $\alpha=2+\sqrt{3\vphantom{\large3}}$ as a parameter and take a derivative w.r.t. $\alpha$ at this point, we get the following:
$$I=\int_0^1\frac{\ln x}{\left(1+x\right)\left(1+x^{-\left(2+\sqrt{3\vphantom{\large3}}\right)}\right)}dx.$$
Is it possible to express the integral $I$ in a closed form?

Here is a partial progress report. I am basically repeating Jim Belk's analysis from the previous answer.  
Set $F(a) = \int_{x=0}^1 \frac{\log(1+x^a)}{1+x} dx$. Then
$$F(a) = \int_{x=0}^1 \int_{y=0}^{x^a} \frac{dx dy}{(1+x)(1+y)} = \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)}$$
so
$$F(a) + F(a^{-1}) = \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)} + \int_{0 \leq y \leq x^{1/a} \leq 1} \frac{dx dy}{(1+x)(1+y)}$$
$$= \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)} + \int_{0 \leq y^a \leq x \leq 1} \frac{dx dy}{(1+x)(1+y)} = \int_{0 \leq x,y \leq 1} \frac{dx dy}{(1+x)(1+y)} = (\log 2)^2.$$
(In order to combine the integrals, first switch the names of $x$ and $y$ in the second one.)
So 
$$F'(a) - a^{-2} F'(a^{-1})=0.$$
This gives a linear relation between $F'(2 + \sqrt{3})$ and $F'(2-\sqrt{3})$. If we find a second one, we can solve the linear equations and be done.

Notice that
$$F'(a) = \int_{x=0}^1 \frac{x^a \log x dx}{(1+x)(1+x^a)} = \sum_{m,n=0}^{\infty} \int_{x=0}^1 (-1)^{m+n} x^{m+(n+1) a} \log x dx.$$
Integrating by parts, $\int_{x=0}^1 x^b \log x dx = \frac{-1}{(b+1)^2}$. So, ignoring issues of convergence, we should have
$$F'(a) =  \sum_{m,n=0}^{\infty}  \frac{(-1)^{m+n+1}}{(m+(n+1) a + 1)^2} = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1}}{(m+n a)^2}$$
In the last step, we turned $m+1$ and $n+1$ into $m$ and $n$ to make things pretty. My guess is that the convergence issues can be dealt with for any $a>0$, but I haven't thought much about it.
So
$$F'(a) + F'(a^{-1}) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left( \frac{(-1)^{m+n+1}}{(m+n a)^2} +\frac{(-1)^{m+n+1}}{(m+n a^{-1})^2} \right).$$
Putting $a=2 + \sqrt{3}$, this is
$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{m+n+1} \frac{2 (m^2+4mn+7n^2)}{(m^2+4mn+n^2)^2} $$
$$= 2 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}  \frac{(-1)^{m+n+1}}{m^2+4mn+n^2} +12  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}  \frac{(-1)^{m+n+1} n^2}{(m^2+4mn+n^2)^2}.$$
Here is where I run out of ideas. The first sum is basically the one at the end of Jim Belk's post, but I have no ideas for the second one.
