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?
 A: Jim Belk's analysis is very impressive. But I'm afraid there is an assumption that $\alpha\ge 0$ being implicitly made in his analysis. Here presents the other half of the answer.
\begin{eqnarray}
F(\alpha)&=&\int_0^1 {\frac{\ln (1+x^\alpha)}{1+x}dx} \\
I(\alpha)&=&\frac{dF}{d\alpha}=\int_0^1 {\frac{\ln x}{(1+x)(1+x^{-\alpha})}dx} \\
&=&\int_0^1 {\frac{\ln x}{1+x}\frac{x^\alpha}{1+x^\alpha}dx} \\
&=&\int_0^1 {\frac{\ln x}{1+x} \left( 1 - \frac{1}{1+x^\alpha} \right) dx }  \\
&=&\int_0^1 {\frac{\ln x}{1+x}dx}-\int_0^1{ \frac{1}{1+x^\alpha} \frac{\ln x}{1+x}dx } \\
&=&-\frac{\pi^2}{12}-I(-\alpha) \\
\end{eqnarray}
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{1}{\ln\pars{x} \over \pars{1 + x}
     \bracks{1 + x^{-\pars{2 + \root{3}}}}}\,\dd x:\ {\large ?}}$

\begin{align}
\mbox{Let's consider}&\quad
{\cal F}\pars{\mu}\equiv
\int_{0}^{1}{\ln\pars{x} \over \pars{1 + x}\pars{1 + x^{-\mu}}}\,\dd x\quad
\mbox{such that}
\\[3mm]&\int_{0}^{1}{\ln\pars{x}\over
\pars{1 + x}
\bracks{1 + x^{-\pars{2 + \root{3}}}}}\,\dd x = {\cal F}\pars{2 + \root{3}}\tag{1}
\end{align}

\begin{align}
\color{#c00000}{{\cal F}\pars{\mu}}
&=\totald{}{\mu}\int_{0}^{1}{\ln\pars{1 + x^{\mu}} \over 1 + x}\,\dd x
=\totald{}{\mu}\int_{0}^{1}\sum_{m = 1}^{\infty}{\pars{-1}^{m + 1} \over m}x^{m\mu}
\sum_{n = 0}^{\infty}\pars{-1}^{n}x^{n}\,\dd x
\\[3mm]&=\totald{}{\mu}\sum_{n = 0}^{\infty}\pars{-1}^{n}
\sum_{m = 1}^{\infty}{\pars{-1}^{m + 1} \over m}\int_{0}^{1}x^{m\mu + n}\,\dd x
=\totald{}{\mu}\sum_{n = 0}^{\infty}\pars{-1}^{n}
\sum_{m = 1}^{\infty}{\pars{-1}^{m + 1} \over m\pars{m\mu + n}}
\\[3mm]&=\totald{}{\mu}\left\lbrace
{1 \over \mu}\sum_{n = 0}^{\infty}\pars{-1}^{n}\times\right.
\\[3mm]&\left.\phantom{\totald{}{\mu}\braces{\,\,\,}}\bracks{
\sum_{m = 0}^{\infty}{1 \over \pars{2m + 1}\pars{2m + 1 + n/\mu}}
-\sum_{m = 0}^{\infty}{1 \over \pars{2m + 2}\pars{2m + 2 + n/\mu}}}\right\rbrace
\\[3mm]&=
{1 \over 4}\,\totald{}{\mu}\braces{{1 \over \mu}
\sum_{n = 0}^{\infty}\pars{-1}^{n}\bracks{
{\Psi\pars{\bracks{1 + n/\mu}/2} - \Psi\pars{1/2} \over n/\bracks{2\mu}}
-
{\Psi\pars{1 + n/\bracks{2\mu}} - \Psi\pars{1} \over n/\bracks{2\mu}}}}
\\[3mm]&=\half\,\totald{}{\mu}
\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n}\bracks{
\Psi\pars{\half + {n \over 2\mu}} - \Psi\pars{1 + {n \over 2\mu}}}
\\[3mm]&=\color{#c00000}{-\,{1 \over 8\mu^{2}}
\sum_{n = 0}^{\infty}\pars{-1}^{n}\bracks{
-\Psi'\pars{\half + {n \over 2\mu}} + \Psi'\pars{1 + {n \over 2\mu}}}}
\end{align}
where $\ds{\Psi\pars{z}}$ is the
Digamma Function.

Also,
  \begin{align}
\Psi'\pars{\half + {n \over 2\mu}}&=4\Psi'\pars{n \over \mu}
-\Psi'\pars{n \over 2\mu}
\\[3mm]
\Psi'\pars{1 + {n \over 2\mu}}&=\Psi'\pars{n \over 2\mu} - {4\mu^{2} \over n^{2}}
\end{align}

\begin{align}
\color{#00f}{\large{\cal F}\pars{\mu}}&=
{\Psi'\pars{1/2} - \Psi'\pars{1} \over 8\mu^{2}}
+{1 \over 4\mu^{2}}
\sum_{n = 1}^{\infty}\pars{-1}^{n}\bracks{
2\Psi'\pars{n \over \mu} - \Psi'\pars{n \over 2\mu}}
-\half\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2}}
\\[3mm]&=\color{#00f}{\large{\pi^{2} \over 24}\pars{{1 \over \mu^{2}} + 1}
+{1 \over 4\mu^{2}}
\sum_{n = 1}^{\infty}\pars{-1}^{n}\bracks{
2\Psi'\pars{n \over \mu} - \Psi'\pars{n \over 2\mu}}}
\end{align}

So far !!!.

A: 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.
