TITLE: Integral $\int_0^1\frac{\ln x}{\left(1+x\right)\left(1+x^{-\left(2+\sqrt3\right)}\right)}dx$
QUESTION [33 upvotes]: 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?

REPLY [4 votes]: $\newcommand{\+}{^{\dagger}}
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$\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 !!!.