I am in the process of proving $$I=\int_0^\infty \frac{\arctan x}{x^4+x^2+1}\mathrm{d}x=\frac{\pi^2}{8\sqrt{3}}-\frac23G+\frac\pi{12}\log(2+\sqrt{3})$$ And I have gotten as far as showing that $$2I=\frac{\pi^2}{4\sqrt{3}}+J$$ Where $$J=\int_0^\infty \log\bigg(\frac{x^2-x+1}{x^2+x+1}\bigg)\frac{\mathrm{d}x}{1+x^2}$$ Then we preform $x=\tan u$ to see that $$J=\int_0^{\pi/2}\log\bigg(\frac{2+\sin2x}{2-\sin2x}\bigg)\mathrm dx$$ Which I have been stuck on for the past while. I tried defining $$k(a)=\int_0^{\pi/2}\log(2+\sin2ax)\mathrm dx$$ Which gives $$J=k(1)-k(-1)$$ Then differentiating under the integral: $$k'(a)=2\int_0^{\pi/2}\frac{x\cos2ax}{2+\sin2ax}\mathrm dx$$ We may integrate by parts with $u=x$ to get a differential equation $$ak'(a)+k(a)=\frac\pi2\log(2+\sin\pi a)$$ With initial condition $$k(0)=\frac\pi2\log2$$ And from here I have no idea what to do.
I also tried tangent half angle substitution, but that just gave me the original expression for $J$.
I'm hoping that there is some really easy method that just never occurred to me... Any tips?
Edit
As was pointed out in the comments, I could consider $$P(a)=\frac12\int_0^\pi \log(a+\sin x)\mathrm dx\\\Rightarrow P(0)=-\frac\pi2\log2$$ And $$ \begin{align} Q(a)=&\frac12\int_0^\pi \log(a-\sin x)\mathrm dx\\ =&\frac12\int_0^\pi\log[-(-a+\sin x)]\mathrm dx\\ =&\frac12\int_0^\pi\bigg(\log(-1)+\log(-a+\sin x)\bigg)\mathrm dx\\ =&\frac{i\pi}2\int_0^\pi\mathrm{d}x+\frac12\int_0^\pi\log(-a+\sin x)\mathrm dx\\ =&\frac{i\pi^2}2+P(-a) \end{align} $$ Hence $$J=P(2)-Q(2)=P(2)-P(-2)-\frac{i\pi^2}2$$ So now we care about $P(a)$. Differentiating under the integral, we have $$P'(a)=\frac12\int_0^\pi \frac{\mathrm{d}x}{a+\sin x}$$ With a healthy dose of Tangent half angle substitution, $$P'(a)=\int_0^\infty \frac{\mathrm{d}x}{ax^2+2x+a}$$ completing the square, we have $$P'(a)=\int_0^\infty \frac{\mathrm{d}x}{a(x+\frac1a)^2+g}$$ Where $g=a-\frac1a$. With the right trigonometric substitution, $$P'(a)=\frac1{\sqrt{a^2+1}}\int_{x_1}^{\pi/2}\mathrm{d}x$$ Where $x_1=\arctan\frac1{\sqrt{a^2+1}}$. Then using $$\arctan\frac1x=\frac\pi2-\arctan x$$ We have that $$P'(a)=\frac1{\sqrt{a^2+1}}\arctan\sqrt{a^2+1}$$ So we end up with something I don't know how to to deal with (what a surprise) $$P(a)=\int\arctan\sqrt{a^2+1}\frac{\mathrm{d}a}{\sqrt{a^2+1}}$$ Could you help me out with this last one? Thanks.