Evaluate $\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx$ I am trying to find a closed form for
$$\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx = 0.094561677526995723016 \cdots$$
It seems that the answer is
$$\frac{\pi^2}{12}\left( 1-\sqrt{3}\right)+\log(2) \log \left(1+\sqrt{3} \right)$$
Mathematica is unable to give a closed form for the indefinite integral.
How can we prove this result? Please help me.
EDIT
Apart from this result, the following equalities are also known to exist:
$$\begin{align*}
\int_0^1 \frac{\log \left( 1+x^{4+\sqrt{15}}\right)}{1+x}\mathrm dx &=\frac{\pi^2}{12} \left( 2-\sqrt{15}\right)+\log \left( \frac{1+\sqrt{5}}{2}\right)\log \left(2+\sqrt{3} \right) \\ &\quad +\log(2)\log\left( \sqrt{3}+\sqrt{5}\right)
\\ \int_0^1 \frac{\log \left( 1+x^{6+\sqrt{35}}\right)}{1+x}\mathrm dx &= \frac{\pi^2}{12} \left( 3-\sqrt{35}\right)+\log \left(\frac{1+\sqrt{5}}{2} \right)\log \left(8+3\sqrt{7} \right) \\
&\quad +\log(2) \log \left( \sqrt{5}+\sqrt{7}\right)
\end{align*}$$
Please take a look here.
 A: I have found an interesting reference related to this integral. It may be of interest to many users:
A restricted Epstein zeta function
and the evaluation of some definite integrals by Habib Muzaffar and Kenneth S. Williams 
A: This isn't an answer to the question, but I thought I should post some of my work here.  
Consider the function
$$
F(a) \;=\; \int_0^1 \frac{\log(1+x^a)}{1+x}dx.
$$
The question asks us to prove that $F(2+\sqrt{3}) = \dfrac{\pi^2}{12}(1+\sqrt{3}) + \log(2)\log(1+\sqrt{3})$.


*

*Mathematica isn't able to compute a closed form for $F(2+\sqrt{3})$, but it can compute $F(a)$ for certain other values of $a$:


*

*$F(0) \;=\; (\log 2)^2.$  (Very easy.)

*$F(1) \;=\; \dfrac{1}{2}(\log 2)^2.$  (Also easy to do by hand.)

*$F(2) \;=\; -\dfrac{\pi^2}{48} \,+\, \dfrac{3}{4}(\log 2)^2$

*$F(3) \;=\; \dfrac{\pi^2}{9}+\dfrac{1}{2}(\log 2)^2 \,+\, \dfrac{1}{2}\mathrm{Li}_2\left(-\dfrac13\right) - \mathrm{Li}_2\left(\dfrac{1+i\sqrt{3}}6\right) - \mathrm{Li}_2\left(\dfrac{1-i\sqrt{3}}6\right)$
It can also compute $F(1/3)$, $F(1/2)$, $F(-2)$, $F(-1)$, $F(-1/2)$, and $F(-1/3)$, which follow from the symmetry properties below, as well as very long expressions for $F(4)$, $F(-4)$, $F(1/4)$, and $F(-1/4)$.
In the formula for $F(3)$, the function $\mathrm{Li}_2$ is the dilogarithm, which is defined by the integral
$$
\mathrm{Li}_2(a) \;=\; -\int_0^1 \frac{\ln(1-ax)}{x}dx.
$$
Interestingly, the only values of $a$ for which $\mathrm{Li}_2(a)$ is known to have a closed form are $-1$, $0$, $1/2$, $1$, $2$, and various expressions involving $\sqrt{5}$. (See here.)

*The function $F(a)$ defined above has a few symmetry properties.  It is easy to show that
$$
F(-a) \;=\; F(a) - a\int_0^1 \frac{\log x}{1+x}dx.
$$
which gives
$$
F(-a) \;=\; F(a) - \frac{\pi^2 a}{12}.
$$
In addition, integration by parts followed by a substitution can be used to show that
$$
F(a) \,+\, F(1/a) \;=\; (\log 2)^2
$$
for any positive value of $a$.  Since $(2+\sqrt{3})^{-1} = 2-\sqrt{3}$, it follows that the given question is equivalent to the equation
$$
F(2-\sqrt{3}) \;=\; \frac{\pi^2}{12}(\sqrt{3}-1)\,+\,\log(2)\log(\sqrt{3}-1).
$$

*It's not too hard to find a series for $F(\alpha)$.  We have
$$
\begin{align*}
\frac{\log(1+x^\alpha)}{1+x} \;&=\; \left(\frac{1}{1+x}\right)\log(1+x^\alpha) \\[6pt]
&=\; \left(\sum_{k=1}^\infty (-1)^{k+1}x^{k-1}\right)\sum_{n=1}^\infty (-1)^{n+1}\frac{x^{\alpha n}}{n} \\[6pt]
&=\; \sum_{n,k=1}^\infty (-1)^{n+k}\frac{x^{\alpha n+k-1}}{n}
\end{align*}
$$
We can now (by Abel's Theorem) integrate this series term-by-term to get
$$
F(\alpha) \;=\; \sum_{n,k=1}^\infty \frac{(-1)^{n+k}}{n(\alpha n+k)}
$$
In particular
$$
F(2+\sqrt{3}) \;=\; \sum_{n,k=1}^\infty \frac{(-1)^{n+k}}{n\bigl((2+\sqrt{3})n+k\bigr)}
$$
The $2+\sqrt{3}$ in the denominator of this series is unfortunate.  However, we already know the value of the sum $F(2+\sqrt3)+F(2-\sqrt3)$, so all we really want is the value of the difference $F(2+\sqrt3)-F(2-\sqrt3)$.  In general:
$$
\begin{align*}
F(\alpha)-F(\alpha^{-1}) \;&=\; \sum_{n,k=1}^\infty (-1)^{n+k}\left(\frac{1}{n(\alpha n+k)}-\frac{1}{n(\alpha^{-1} n+k)}\right) \\[6pt]
&=\; \sum_{n,k=1}^\infty (-1)^{n+k}\frac{(\alpha^{-1} n+k)-(\alpha n+k)}{n(\alpha n+k)(\alpha^{-1} n+k)} \\[6pt]
&=\; \bigl(\alpha^{-1}-\alpha\bigr)\sum_{n,k=1}^\infty \frac{(-1)^{n+k}}{n^2 + (\alpha+\alpha^{-1})nk + k^2}
\end{align*}
$$
So:
$$
F(2+\sqrt{3})-F(2-\sqrt{3}) \;=\; -2\sqrt{3} \sum_{n,k=1}^\infty \frac{(-1)^{n+k}}{n^2 + 4nk + k^2}
$$
This series is more pleasant, but I still have no idea how to evaluate it.
A: Unfortunately, the following generalization works for only positive integer $a$:
$$\int_0^1\frac{\ln(1+x^{2a})}{1+x}dx=\ln^2(2)-\frac{2a^2-1}{8a}\zeta(2)+\frac12\sum_{j=0}^{2a-1}\ln^2\left(2\sin\left(\frac{(2j+1)\pi}{4a}\right)\right)$$
$$-\frac12\sum_{j=0}^{a-1}\ln^2\left(2\sin\left(\frac{(2j+1)\pi}{2a}\right)\right).$$
Proof:
$$\int_0^1\frac{\ln(1+x^{2a})}{1+x}dx=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\int_0^1\frac{x^{2an}}{1+x}dx$$
$$=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(\ln(2)+H_{an}-H_{2an}\right)$$
$$=\ln^2(2)+\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(H_{an}-H_{2an}\right).$$
Omram Kouba provided in his paper page (12-13) the following general result:
$$\sum_{n=1}^\infty(-1)^{n-1}\frac{H_{an}}{n} = \frac{a^2+1}{4a}\zeta(2) - \frac{1}{2} \sum_{j=0}^{a-1} \ln^2\left(2 \sin \frac{(2j+1)\pi}{2a} \right)$$
from which, the proof follows.
By the way, we can find the integral representation of $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}H_{an}}{n}$:
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_{an}}{n}=a\sum_{n=1}^\infty (-1)^{n-1}\left(\frac{H_{an}}{an}\right)$$
$$=a\sum_{n=1}^\infty (-1)^{n-1}\left(-\int_0^1 x^{an-1}\ln(1-x)dx\right)$$
$$=a\int_0^1\frac{\ln(1-x)}{x}\left(\sum_{n=1}^\infty(-x^a)^n\right)dx$$
$$=a\int_0^1\frac{\ln(1-x)}{x}\left(\frac{-x^a}{1+x^a}\right)dx$$
$$=a\int_0^1\frac{\ln(1-x)}{x}\left(-1+\frac{1}{1+x^a}\right)dx$$
$$=a\zeta(2)+a\int_0^1\frac{\ln(1-x)}{x(1+x^a)}dx.$$
Substitute the result of $\displaystyle\sum_{n=1}^\infty\frac{(-1)^nH_{an}}{n}$, we also get
$$\int_0^1\frac{\ln(1-x)}{x(1+x^a)}dx=\frac{1-3a^2}{4a^2}\zeta(2)- \frac{1}{2a} \sum_{j=0}^{a-1} \ln^2\left(2 \sin \frac{(2j+1)\pi}{2a} \right).$$
