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.
from Hot Weekly Questions - Mathematics Stack Exchange
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