Solving the integral $\int_0^{\pi/2}\log\left(\frac{2+\sin2x}{2-\sin2x}\right)\mathrm dx$ 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.
 A: Result
I find that the integral has a closed form given by
$$i = \int\limits_0^{\pi/2}\log\bigg(\frac{2+\sin2x}{2-\sin2x}\bigg)\mathrm dx = \frac{1}{3} \left(8 C-\pi  \log \left(2+\sqrt{3}\right)\right) \simeq 1.06346\tag{1}$$ 
where 
$$C = \sum _{k=1}^{\infty } \frac{(-1)^{k+1}}{(2 k-1)^2} \simeq 0.915966$$
is Catalan's constant.
Heuristic derivation
Notice trivially that because of the symmtery of the integrand the integral can be written as twice the integral from $0$ to $\frac{\pi}{4}$ which we shall utilize in what follows.
The basic idea is the series expansion
$$\log \left(\frac{1+z}{1-z}\right)=2\tanh ^{-1}(z) = 2 \sum _{k=1}^{\infty } \frac{z^{2 k-1}}{2 k-1},|z|<1 \tag{2}$$
The integral is then to be done over the odd powers of the $\sin$ with the  result 
$$\int_0^{\frac{\pi }{4}} \sin ^{2 k-1}(2 x) \, dx = \frac{\sqrt{\pi } \Gamma (k)}{4 \Gamma \left(k+\frac{1}{2}\right)}\tag{3}$$
Assembling the pieces the sum to be taken to represent $i$ becomes
$$i_s = \sum _{k=1}^{\infty } \frac{\sqrt{\pi } \Gamma (k)}{(2 k-1) 2^{2 k-1} \Gamma \left(k+\frac{1}{2}\right)}\tag{4}$$
and this sum is immediately computed by Mathematica to give the compact result $(1)$.
Let us make the sum more transparent using the chain
$$\frac{\sqrt{\pi } \Gamma (k)}{\Gamma \left(k+\frac{1}{2}\right)}=B\left(\frac{1}{2},k\right)=\int_0^1 \frac{t^{k-1}}{\sqrt{1-t}} \, dt\tag{5}$$
and doing the sum under the integral
$$\sum _{k=1}^{\infty } \frac{t^{k-1}}{(2 k-1) 2^{2 k-1}}=\frac{\tanh ^{-1}\left(\frac{\sqrt{t}}{2}\right)}{\sqrt{t}}\tag{6}$$
leads finally to the integral
$$\int_0^1 \frac{\tanh ^{-1}\left(\frac{\sqrt{t}}{2}\right)}{\sqrt{t} \sqrt{1-t}} \, dt\tag{7}$$
for which Mathematica again quickly gives (1).
But there must be a shorter way ... yes, it is, substituting $\sin (2 x)=\sqrt{t}$ in the original integral gives (7) directly.
A: A Possible way:  Consider
$$I(a)=\int_{0}^{+\infty}\frac{\arctan(ax)}{1+x^2+x^4} dx$$
and 
$$I'(a)=\int_{0}^{+\infty}\frac{x}{(1+x^2+x^4)(1+x^2a^2)}dx=\int_{0}^{+\infty}\frac{1}{(1+y+y^2)(1+a^2y)}dx$$
and
$$\frac{1}{(1+y+y^2)(1+a^2y)}= \frac{-a^2y-a^2+1}{(a^4-a^2+1)(1+y+y^2)}+\frac{a^4}{(a^4-a^2+1)(ay^2+1)}$$
we can also calculate $I'(a)$ by complex integration (if you've learned that).
Thanks to Dylan for his advice.
A: $$J=\int_0^{\pi/2}\ln\left(\frac{2+\sin2x}{2-\sin2x}\right)\mathrm dx\overset{2x=t}=\frac12 \int_0^\pi \ln\left(\frac{1+\frac12\sin t}{1-\frac12\sin t}\right)\mathrm dt=\int_0^\frac{\pi}{2}\ln\left(\frac{1+\frac12\sin x}{1-\frac12\sin x }\right)\mathrm dx$$
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: $$2I=\frac{\pi^2}{4\sqrt 3}- \int_0^\infty\frac{(x^2-1)\arctan x}{x^4+x^2+1}dx=\frac{\pi^2}{4\sqrt 3}-\frac12\underbrace{\int_0^\infty \ln\bigg(\frac{x^2-x+1}{x^2+x+1}\bigg)\frac{dx}{1+x^2}}_{=J}$$
A: $$\begin{align*}
\mathcal J &= \int_0^\infty \log\left(\frac{x^2-x+1}{x^2+x+1}\right)\frac{dx}{1+x^2} \\[1ex]
&= \int_0^1 \log\left(\frac{x^2-x+1}{x^2+x+1}\right)\frac{dx}{1+x^2} + \int_1^\infty \log\left(\frac{x^2-x+1}{x^2+x+1}\right)\frac{dx}{1+x^2} \\[1ex]
&= 2 \int_0^1 \log\left(\frac{x^2-x+1}{x^2+x+1}\right)\frac{dx}{1+x^2} & x\mapsto\frac1x\\[1ex]
&= 2 \int_0^1 \ln \left(\frac{1+3x^2}{3+x^2}\right) \frac{dx}{1+x^2} & x\mapsto\frac{1-x}{1+x} \\[1ex]
&= 2 \underbrace{\int_0^1 \frac{\ln(1+3x^2)}{1+x^2} \, dx}_{J_1} - 2 \underbrace{\int_0^1 \frac{\ln(3+x^2)}{1+x^2} \, dx}_{J_2}
\end{align*}$$
and use $J_1 = I(3)$ and $J_2 = I(3)$.
A: $$
\begin{aligned}
J &=\int_0^{\frac{\pi}{2}} \ln \left(\frac{2+\sin 2 x}{2-\sin 2 x}\right) d x \\
&=\frac{1}{2} \left[\int_0^\pi \ln (2+\sin x) d x-\int_0^\pi \ln (2-\sin x) d x\right]
\end{aligned}
$$
By my post, we have
$$
\begin{aligned}
J=& \frac{1}{2}\left[-\pi \ln 2+\frac{2 \pi}{3} \ln (2+\sqrt{3})+\frac{8}{3} G-\left(-\pi \ln 2+\frac{4 \pi}{3} \ln (2+\sqrt{3})-\frac{8}{3} G\right)\right] \\
=&-\frac{\pi}{3} \ln (2+\sqrt{3})+\frac{8}{3} G
\end{aligned}
$$
