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.
Best Answer
$$J=\int_0^{\pi/2}\ln\left(\frac{2+\sin2x}{2-\sin2x}\right)\mathrm dx\overset{2x=t}=\frac12 \int_0^\pi \ln\left(\frac{1+\frac12\sin t}{1-\frac12\sin t}\right)\mathrm dt=\int_0^\frac{\pi}{2}\ln\left(\frac{1+\frac12\sin x}{1-\frac12\sin x }\right)\mathrm dx$$ Now let's consider the following integral: $$I(a)=\int_0^\frac{\pi}{2}\ln\left(\frac{1+\sin a\sin x}{1-\sin a\sin x}\right)dx\Rightarrow I'(a)=2\int_0^\frac{\pi}{2} \frac{\cos a\sin x}{1-\sin^2a\sin^2 x}dx$$ $$=\frac{2\cos a}{\sin^2a}\int_0^\frac{\pi}{2} \frac{\sin x}{\cos^2x +\cot^2 a}dx=\frac{2}{\sin a}\,\arctan\left(x\tan a\right)\bigg|_0^1=\frac{2a}{\sin a}$$ $$I(0)=0 \Rightarrow J=I\left(\frac{\pi}{6}\right)=2\int_0^\frac{\pi}{6}\frac{x}{\sin x}dx$$ $$=2\int_0^{\frac{\pi}{6}} x \left(\ln\left(\tan \frac{x}{2}\right)\right)'dx=2x \ln\left(\tan \frac{x}{2}\right)\bigg|_0^{\frac{\pi}{6}} -2{\int_0^{\frac{\pi}{6}} \ln\left(\tan \frac{x}{2}\right)dx}=$$ $$\overset{\frac{x}{2}=t}=\frac{\pi}{3}\ln(2-\sqrt 3) -4\int_0^\frac{\pi}{12}\ln (\tan t)dt=\frac{\pi}{3}\ln(2-\sqrt 3) +\frac{8}{3}G$$ $G$ is the Catalan's constant and for the last integral see here.
Also note that there's a small mistake. After integrating by parts you should have: $$2I=\frac{\pi^2}{4\sqrt 3}- \int_0^\infty\frac{(x^2-1)\arctan x}{x^4+x^2+1}dx=\frac{\pi^2}{4\sqrt 3}-\frac12\underbrace{\int_0^\infty \ln\bigg(\frac{x^2-x+1}{x^2+x+1}\bigg)\frac{dx}{1+x^2}}_{=J}$$