If $n$ is such that $2n−1$ and $2n+1$ are twin primes,
$$∫_1^0\frac{\log(1+x^{2n+\sqrt{4n^2−1})}}{1+x}dx$$
appears to be
$$\frac{\pi^{2}}{12}\left(n-\sqrt{4n^{2}-1}\right)+\log(a+b\sqrt{2n-1})\log(c+d \sqrt{2n+1})+\log(2)\log(\sqrt{2n-1}+\sqrt{2n+1})$$
for some integers $a,b,c,d$ with $a^2-(2n-1) b^2 = -1,c^2-(2n+1)d^2=1$
Examples :
$$\begin{align*}\int_0^1 \frac{\log \left( 1+x^{4+\sqrt{15}}\right)}{1+x}\mathrm dx &=\frac{\pi^2}{12} \left( 2-\sqrt{15}\right)+\log \left( \frac{1+\sqrt{5}}{2}\right)\log \left(2+\sqrt{3} \right) \\ &\quad +\log(2)\log\left( \sqrt{3}+\sqrt{5}\right)\\ \int_0^1 \frac{\log \left( 1+x^{6+\sqrt{35}}\right)}{1+x}\mathrm dx &= \frac{\pi^2}{12} \left( 3-\sqrt{35}\right)+\log \left(\frac{1+\sqrt{5}}{2} \right)\log \left(8+3\sqrt{7} \right) \\&\quad +\log(2) \log \left( \sqrt{5}+\sqrt{7}\right)\end{align*}$$.
Why ??
for strict positive integer $a$ we have :
$$\int_0^1\frac{\ln(1+x^{2a})}{1+x}dx=\ln^2(2)-\frac{2a^2-1}{8a}\zeta(2)+\frac12\sum_{j=0}^{2a-1}\ln^2\left(2\sin\left(\frac{(2j+1)\pi}{4a}\right)\right)$$$$-\frac12\sum_{j=0}^{a-1}\ln^2\left(2\sin\left(\frac{(2j+1)\pi}{2a}\right)\right).$$