Integral $\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx$ 
Prove that$$I=\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G$$

I've found this integral in my notebook and perhaps I encountered it before since it looks quite familiar.
Anyway I thought it's quite a trivial integral so I'm gonna solve it quickly, but I am having some hard time to finish it. I went on with Feynman's trick:
$$I(a)=\int_0^\infty \frac{\ln((1+x^2)a+x)}{1+x^2}dx\Rightarrow I'(a)=\int_0^\infty \frac{dx}{a+x+ax^2}$$
$$=\frac1a\int_0^\infty \frac{dx}{\left(x+\frac{1}{2a}\right)^2+1-\frac{1}{4a^2}}=\frac{1}{a}\frac{1}{\sqrt{1-\frac{1}{4a^2}}}\arctan\left(\frac{x+\frac{1}{2a}}{\sqrt{1-\frac{1}{4a^2}}}\right)\bigg|_0^\infty$$$$=\frac{\pi}{\sqrt{4a^2-1}}-\frac{2}{\sqrt{4a^2-1}}\arctan\left(\frac{1}{\sqrt{4a^2-1}}\right)=\frac{2\arctan\left(\sqrt{4a^2-1}\right)}{\sqrt{4a^2-1}}$$
We can prove easily via the substitution $x\to \frac{1}{x}$ that $I(0)=0$ so we have that:
$$I=I(1)-I(0)=2\int_0^1 \frac{\arctan\left(\sqrt{4a^2-1}\right)}{\sqrt{4a^2-1}}da$$
Now I thought about two substitutions:
$$ \overset{a=\frac12\cosh x}=\int_{\operatorname{arccosh}(0)}^{\operatorname{arccosh}(2)} \arctan(\sinh x)dx$$
$$\overset{a=\frac12\sec x}=\int_{\operatorname{arcsec}(0)}^{\frac{\pi}{3}}\frac{x}{\cos x}dx$$
But in both cases the lower bound is annoying and I think I am missing something here (maybe obvious).
So I would love to get some help in order to finish this.

Edit: We can apply once again Feynman's trick. First consider: $$I(t)=\int_0^1 \frac{2\arctan(t\sqrt{4a^2-1})}{\sqrt{4a^2-1}}da\Rightarrow I'(t)=2\int_0^1 \frac{1}{1+t^2(4a^2-1)}da$$
$$=\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2at}{\sqrt{1-t^2}}\right)\bigg|_0^1=\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2t}{\sqrt{1-t^2}}\right)$$
So once again we have $I(0)=0$, so $I=I(1)-I(0)$.
$$\Rightarrow I=\int_0^1\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2t}{\sqrt{1-t^2}}\right)dt\overset{t=\sin x}=\int_0^\frac{\pi}{2}\frac{\arctan(2\tan x)}{\sin x}dx$$
At this point Mathematica can evaluate the integral to be:
$$I=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G$$
I didn't try the last integral yet, but I am thinking of Feynman again $\ddot \smile$.

Edit 2: Found that I already was on it some time ago, and actually posted it here, which means I have solved it before using Feynman's trick, but right now I can't remember how I did it.
So given the circumstances I am positive that it can be solved starting with my approach, but if you have any other ways then feel free to share it.

Solution 3. Consider the following integral:
$$ I(a)=\int_0^\frac{\pi}{2}\frac{\arctan(a\tan x)}{\sin x}dx\Rightarrow I'(a)=\int_0^\frac{\pi}{2}\frac{\sec x}{1+a^2\tan^2 x}dx$$
$$ =\int_0^\frac{\pi}{2}\frac{\cos x}{\cos^2 x+a^2\sin^2 x}dx\overset{\sin x=y}=\int_0^1 \frac{dy}{1+(a^2-1)y^2}=\frac{\arctan\sqrt{a^2-1}}{\sqrt{a^2-1}}$$
$$\rm I=\underbrace{I(2)-I(1)}_{=J}+I(1), \quad  I(1)=\int_0^\frac{\pi}{2}\frac{x}{\sin x}dx$$
$$J=\int_1^2 \frac{\arctan\sqrt{a^2-1}}{\sqrt{a^2-1}}da\overset{a=\sec x}=\int_0^\frac{\pi}{3}\frac{x}{\cos x}dx\overset{x=\frac{\pi}{2}-t}=\int_\frac{\pi}{6}^\frac{\pi}{2}\frac{\frac{\pi}{2}-t}{\sin t}dt$$
$$\rm=\frac{\pi}{2}\int_\frac{\pi}{6}^\frac{\pi}{2} \frac{1}{\sin t}dt- \int_0^\frac{\pi}{2} \frac{t}{\sin t}dt+\int_0^\frac{\pi}{6} \frac{t}{\sin t}dt$$
$$ \Rightarrow I=J+I(1)=\frac{\pi}{2}\ln\left(\tan \frac{x}{2}\right)\bigg|_\frac{\pi}{6}^\frac{\pi}{2}+\int_0^\frac{\pi}{6} \frac{t}{\sin t}dt$$
Finally, using the result from here, we get:
$$ I=\frac{\pi}{2}\ln(2+\sqrt 3)-\frac{\pi}{6}\ln(2+\sqrt 3)+\frac43G=\boxed{\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G}$$

Solution 4.
$$I=\int_0^2\frac{\arctan\sqrt{a^2-1}}{\sqrt{a^2-1}}da=\int_0^2\frac{\operatorname{arcsec} a}{\sqrt{a^2-1}}da$$
$$\sf I=\int_0^1\frac{\operatorname{arcsec} a}{\sqrt{a^2-1}}da+\int_1^2\frac{\operatorname{arcsec} a}{\sqrt{a^2-1}}da$$
The second integral is like the one from above, and for the first integral we need to use the complex definition of $\sf \arccos z$, namely $\sf -i\ln\left(z+\sqrt{z^2-1}\right)$.
$$\sf \Rightarrow \frac{\operatorname{arcsec} a}{\sqrt{a^2-1}}=\frac{-\ln\left(\frac{1-\sqrt{1-a^2}}{a}\right)}{\sqrt{1-a^2}}$$
And now via the substitution $a=\sin y$ everything goes smoothly.
