一个有意思的对数积分的一般形式
\[\Large\displaystyle\int_0^\infty\frac{\ln\left(\displaystyle\frac{1+x^{4+\sqrt{15\vphantom{\large A}}}}{1+x^{2+\sqrt{3\vphantom{\large A}}}}\right)}{\left(1+x^2\right)\ln x}\mathrm dx=\frac{\pi}{4}\left(2+\sqrt{6}\sqrt{3-\sqrt{5}}\right)
\]
\(\Large\mathbf{Proof:}\)
\(\large\mathbf{Method ~One:}\)
Assume \(a,b\in\mathbb{R}\). Note that
\[\int_0^\infty\frac{\ln\left(\displaystyle\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{\mathrm dx}{1+x^2}=\int_0^\infty\frac{\ln\left(\displaystyle\frac{1+x^a}2\right)}{\ln x}\frac{\mathrm dx}{1+x^2}-\int_0^\infty\frac{\ln\left(\displaystyle\frac{1+x^b}2\right)}{\ln x}\frac{\mathrm dx}{1+x^2}.
\]
Both integrals on the right-hand side have the same shape, so we only need to evaluate one of them:
\[\begin{align*}
&\phantom=\underbrace{\int_0^\infty\frac{\ln\left(\displaystyle\frac{1+x^a}2\right)}{\ln x}\frac{\mathrm dx}{1+x^2}}_\text{split the region}\\
&=\int_0^1\frac{\ln\left(\displaystyle\frac{1+x^a}2\right)}{\ln x}\frac{\mathrm dx}{1+x^2}+\underbrace{\int_1^\infty\frac{\ln\left(\displaystyle\frac{1+x^a}2\right)}{\ln x}\frac{\mathrm dx}{1+x^2}}_{\text{change variable}\ y=1/x}\\
&=\int_0^1\frac{\ln\left(\displaystyle\frac{1+x^a}2\right)}{\ln x}\frac{\mathrm dx}{1+x^2}+\underbrace{\int_1^0\frac{\ln\left(\displaystyle\frac{1+y^{-a}}2\right)}{\ln\left(y^{-1}\right)}\frac1{1+y^{-2}}\left(-\frac1{y^2}\right)\mathrm dy}_\text{flip the bounds and simplify}\\
&=\int_0^1\frac{\ln\left(\displaystyle\frac{1+x^a}2\right)}{\ln x}\frac{\mathrm dx}{1+x^2}-\underbrace{\int_0^1\frac{\ln\left(\displaystyle\frac{1+y^{-a}}2\right)}{\ln y}\frac{\mathrm dy}{1+y^2}}_{\text{rename}\ y\ \text{to}\ x}\\
&=\underbrace{\int_0^1\frac{\ln\left(\displaystyle\frac{1+x^a}2\right)}{\ln x}\frac{\mathrm dx}{1+x^2}-\int_0^1\frac{\ln\left(\displaystyle\frac{1+x^{-a}}2\right)}{\ln x}\frac{\mathrm dx}{1+x^2}}_\text{combine logarithms}\\
&=\int_0^1\frac{\ln\left(\displaystyle\frac{1+x^a}{1+x^{-a}}\right)}{\ln x}\frac{\mathrm dx}{1+x^2}=\underbrace{\int_0^1\frac{\ln\left(\displaystyle\frac{x^a\left(x^{-a}+1\right)}{1+x^{-a}}\right)}{\ln x}\frac{\mathrm dx}{1+x^2}}_{\text{cancel}\ \ 1+x^{-a}}\\
&=\int_0^1\frac{\ln\left(x^a\right)}{\ln x}\frac{\mathrm dx}{1+x^2}=a\int_0^1\frac{\mathrm dx}{1+x^2}=a\,\Big(\arctan1-\arctan0\Big)\\&=\vphantom{\Bigg|^0}\frac{\pi\,a}4
\end{align*}\]
So, finally,
\[\int_0^\infty\frac{\ln\left(\displaystyle\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{\mathrm dx}{1+x^2}=\frac\pi4(a-b)
\]
Therefore,
\[\Large\boxed{\displaystyle\int_0^\infty\frac{\ln\left(\displaystyle\frac{1+x^{4+\sqrt{15\vphantom{\large A}}}}{1+x^{2+\sqrt{3\vphantom{\large A}}}}\right)}{\left(1+x^2\right)\ln x}\mathrm dx=\color{Blue} {\dfrac{\pi}{4}(2+\sqrt{15}-\sqrt{3})}}
\]
\(\large\mathbf{Method ~Two:}\)
Let the considered integral be \(I\). Just to make it easier to write, let \(4+\sqrt{15}=a\) and \(2+\sqrt{3}=b\). Use the substitution \(x=\tan\theta\) to get:
\[I=\int_0^{\pi/2} \frac{\ln\left(\dfrac{1+(\tan\theta)^a}{1+(\tan\theta)^b}\right)}{\ln\tan\theta}\,\mathrm d\theta
\]
Next, use the substitution \(\theta=\pi/2-t\) to obtain:
\[I=\int_0^{\pi/2} \frac{\ln\left(\dfrac{1+(\tan t)^a}{(1+(\tan t)^b)(\tan t)^{a-b}}\right)}{\ln\cot t}\,\mathrm dt=\int_0^{\pi/2} \frac{-\ln\left(\dfrac{1+(\tan\theta)^a}{1+(\tan\theta)^b}\right)+(a-b)\ln\tan\theta}{\ln\tan\theta}\,\mathrm d\theta
\]
where I used \(\ln(\tan(\pi/2-\theta))=\ln(\cot\theta)=-\ln(\tan\theta)\).Add the two expressions for \(I\) and notice that you are left with:
\[2I=\int_0^{\pi/2} \frac{(a-b)\ln \tan\theta}{\ln \tan\theta}\,d\theta=\frac{\pi}{2}(a-b)
\]
\[I=\frac{\pi}{4}(a-b)
\]
Therefore,
\[\Large\boxed{\displaystyle\int_0^\infty\frac{\ln\left(\displaystyle\frac{1+x^{4+\sqrt{15\vphantom{\large A}}}}{1+x^{2+\sqrt{3\vphantom{\large A}}}}\right)}{\left(1+x^2\right)\ln x}\mathrm dx=\color{Blue} {\dfrac{\pi}{4}(2+\sqrt{15}-\sqrt{3})}}
\]
