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.
 A: Start by letting $x\mapsto\tan x$ we obtain
$$\int_0^\infty\frac{\log(1+x+x^2)}{1+x^2}\mathrm dx\stackrel{x\mapsto\tan x}=\int_0^\frac\pi2\log(1+\tan x+\tan^2x)\mathrm dx=\int_0^\frac\pi2\log\left(\frac{1+\sin x\cos x}{\cos^2x}\right)\mathrm dx$$
Splitting the logarithm we are left with a standard integral, solvable by differentiating the Beta Function for instance, and another one which I already referred to within the comments. To be precise we get
\begin{align*}
\int_0^\frac\pi2\log\left(\frac{1+\sin x\cos x}{\cos^2x}\right)\mathrm dx&=\pi\log 2+\int_0^\frac\pi2\log(1+\sin x\cos x)\mathrm dx\\
&=\pi\log 2+2\int_0^\frac\pi4\log\left(1+\frac12\sin2x\right)\mathrm dx\\
&=\pi\log 2+\int_0^\frac\pi2\log\left(1+\frac12\sin x\right)\mathrm dx\\
&=\frac\pi2\log2+\int_0^\frac\pi2\log\left(2+\sin x\right)\mathrm dx
\end{align*}
The latter integral $-$ even a more general case $-$ is examined within this AoPS thread. An expression is deduced by the user gustin33. I won't copy his derivation here since his own solution is impressive enough. For the given case he obtained
$$\int_0^\frac\pi2\log\left(2+\sin x\right)\mathrm dx=\frac{4G}3+\frac\pi3\log(2+\sqrt3)-\frac\pi2\log2 $$
Which overall yields to the result.

$$\therefore~\int_0^\infty\frac{\log(1+x+x^2)}{1+x^2}\mathrm dx~=~\frac{4G}3+\frac\pi3\log(2+\sqrt3)$$

The crucial point of the linked post is the identity
$$\int_0^\frac\pi2\log(a+\sin x)\mathrm dx=2\operatorname{Ti}_2(a+\sqrt{a^2-1})-\frac\pi2(\log2+\cosh^{-1}a)$$
For $a=2$ the result follows. I will see if I can find another proof for this identity; otherwise I will just leave this here.

EDIT I
Maybe I am on the right track now! Using the integral representation for the Dilogarithm used in this post and reexpressing the Inverse Tangent Integral in terms of the Dilogarithm aswell we obtain
$$\small
\begin{align*}
\operatorname{Ti}_2(a+\sqrt{a^2-1})&=\frac1{2i}\left[\operatorname{Li}_2(ia+i\sqrt{a^2-1})-\operatorname{Li}_2(-ia+-i\sqrt{a^2-1})\right]\\
&=\frac1{2i}\left[\int_0^1\frac{ia+i\sqrt{a^2-1}}{(ia+i\sqrt{a^2-1})t-1}\log t\mathrm dt-\int_0^1\frac{-ia+-i\sqrt{a^2-1}}{(-ia+-i\sqrt{a^2-1})t-1}\log t\mathrm dt\right]\\
&=\frac{a+\sqrt{a^2-1}}2\int_0^1\left[\frac1{(-1)+i(a+\sqrt{a^2-1})t}+\frac1{(-1)-i(a+\sqrt{a^2-1})t}\right]\log t\mathrm dt\\
&=-(a+\sqrt{a^2-1})\int_0^1\frac{\log t}{1+(a+\sqrt{a^2-1})^2t^2}\mathrm dt
\end{align*}
$$
Mabye this integral is useful for someone. I will try to find something from which it is useful to me too.

EDIT II
The integral can also be reduced to finding
$$\int_0^1\frac{\arctan t}{t^2+t+1}\frac{1-t^2}{1+t^2}\mathrm dt$$
I am almost certain I have seen this one before aswell. I will search for it.
A: Note $\int_0^\infty \frac{\ln x}{1+x^2}dx=0$ and substitute $x=\tan \frac t2$ to rewrite the integral
\begin{align}
I&=\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx
= \int_0^{\frac\pi2}\ln(1+2\sec t)dt
\end{align}
Let $J(a) = \int_0^{\frac\pi2}\ln(1+\sec a\sec t)dt$
$$J’(a)= \int_0^{\frac\pi2}\frac{\sec a\tan a }{\sec a+\cos t}dt
=a\sec a
$$
$$J(0)= \int_0^{\frac\pi2}[\underset{t\to\frac\pi2-t}{\ln(1+\cos t)}- \ln\cos t]dt
= \int_0^{\frac\pi2}\ln(\sec t+\tan t)dt
$$
Then
\begin{align}
I&= J(\frac\pi3)=J(0)+\int_0^{\frac\pi3} J’(a)da 
=\int_0^{\frac\pi2}\ln (\tan t+\sec t) dt
+ \int_0^{\frac\pi3} a\sec a \>\overset{ibp}{da}\\
&= \frac\pi3 \ln\left(\tan  \frac\pi3 +\sec  \frac\pi3\right)+ \int_{ \frac\pi3} ^{\frac\pi2} 
{\ln(\tan a+\sec a) da} \>\>\>\>\>(a=\frac\pi2-2\theta)\\&= \frac\pi3 \ln(2+\sqrt3)-2 \int^{ \frac\pi{12}}_{0} 
\ln\tan\theta \>d\theta
= \frac\pi3 \ln(2+\sqrt3)+\frac43G
\end{align}
A: 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.
A: One more by Feynman’s Technique
Parameterising our integral
$$I=\int_{0}^{\infty} \frac{\ln \left(1+x+x^{2}\right)}{1+x^{2}} d x$$
by the integral
$$
I(a)=\int_{0}^{\infty} \frac{\ln \left(1+2 x \sin a+x^{2}\right)}{1+x^{2}} d x,
$$
where $a\in [-\frac{\pi}{2}, \frac{\pi}{2}]. $
Differentiating $I(a)$ w.r.t. $a$ yields
$$
\begin{aligned}
I^{\prime}(a) &=\int_{0}^{\infty} \frac{2 x \cos a}{\left(1+x^{2}\right)\left(1+2 x \sin a+x^{2}\right)} d x \\
&=\cot a\int_{0}^{\infty}\left(\frac{1}{1+x^{2}}-\frac{1}{1+2 x \sin a+x^{2}}\right) d x \\
&=\cot a\left[\tan ^{-1} x-\frac{1}{\cos a} \tan ^{-1}\left(\frac{x+\sin a}{\cos a}\right)\right]_{0}^{\infty} \\
&=\cot a\left[\frac{\pi}{2}-\frac{1}{\cos a}\left(\frac{\pi}{2}-a\right)\right]
\end{aligned}
$$
Integrating $I’(a)$ back to $I(a)$, we have
$$
\begin{aligned}
I\left(\frac{\pi}{6}\right)- \underbrace{I(0)}_{=\pi\ln 2}  &=\int_{0}^{\frac{\pi}{6}} \cot a\left[\frac{\pi}{2}-\frac{1}{\cos a}\left(\frac{\pi}{2}-a\right)\right] d a \\
&=\frac{\pi}{2} \underbrace{ \int_{0}^{\frac{\pi}{6}}\left(\cot a-\frac{1}{\sin a}\right) d a}_{=\ln \left(\frac{2+\sqrt{3}}{4}\right)} + \underbrace{\int_{0}^{\frac{\pi}{6}} \frac{a}{\sin a} d a}_{K}
\end{aligned}
$$
$$
\begin{aligned}
K &=\int_{0}^{\frac{\pi}{6}} \frac{a}{\sin a} d a=\int_{0}^{\frac{\pi}{6}} a\, d\left[\ln \left(\tan \frac{a}{2}\right)\right] \\
&=\left[a \ln \left(\tan \frac{a}{2}\right)\right]_{0}^{\frac{\pi}{6}}-\int_{0}^{\frac{\pi}{6}} \ln \left(\tan \frac{a}{2}\right) d a \\
&=\frac{\pi}{6} \ln \left(\tan \frac{\pi}{12}\right)-2 \int_{0}^{\frac{\pi}{12}} \ln (\tan a) d a \\
&=-\frac{\pi}{6} \ln (2+\sqrt{3})+\frac{4}{3} G,
\end{aligned}
$$
where $G$ is the Catalan’s constant and the last integral from post.
Now we can conclude that
$$
\boxed{I=I\left(\frac{\pi}{6}\right)=\frac{\pi}{3} \ln (2+\sqrt{3})+\frac{4}{3} G}
$$
