<s>[INST] 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.
 [/INST] 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
$$\int_0^1\frac{\ln(1-x)}{x(1+x^a)}dx=\frac{1-3a^2}{4a^2}\zeta(2)- \frac{1}{2a} \sum_{j=0}^{a-1} \ln^2\left(2 \sin \frac{(2j+1)\pi}{2a} \right).$$
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