<s>[INST] Prove $3\int_{0}^{1}\frac{x\arctan x}{3x^2+1}\,\mathrm dx -\int_{0}^{1}\frac{x\arctan x}{x^2+3}\,\mathrm dx =\frac23 G-\frac {\pi}{12}\ln(2+\sqrt{3})$ 
Prove that
$$
I
=3\int_{0}^{1}\frac{x\arctan x}{3x^2+1}\,\mathrm dx
-\int_{0}^{1}\frac{x\arctan x}{x^2+3}\,\mathrm dx
=\frac23 G-\frac {\pi}{12}\ln(2+\sqrt{3}).
$$

where, $G$ is catalan's constant
Above two Integrals are a part of a integral which I was trying to solve.
Let $I=3I_{1}-I_{2}$
Attempt:-1
$$I_{1}=\int_{0}^{1}\frac{x\arctan x}{x^2+3}\,\mathrm dx$$
$$\implies I_{1}=\int_{0}^{1}\frac{x}{x^2+3}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1} x^{2n-1}\,\mathrm dx$$
$$\implies I_{1}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}\int_{0}^{1}\frac{x^{2n}}{x^2+3}\,\mathrm dx.$$
From my previous question 1 we have
$$
\int_{0}^{1}\frac{x^{2n}}{x^{2}+3}\,\mathrm dx
=(-3)^{n}\frac{\pi}{6\sqrt{3}}+\sum_{k=0}^{n-1}\frac{(-3)^{n-1-k}}{2k+1}.
$$
Therefore,
$$
I_{1}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}\bigg[(-3)^{n}\frac{\pi}{6\sqrt{3}}+\sum_{k=0}^{n-1}\frac{(-3)^{n-1-k}}{2k+1}\bigg],
$$
$$
I_{1}=\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}\frac{(-1)^{2n-k}\space 3^{n-1-k}}{(2k+1)(2n-1)}-\frac{\pi}{6\sqrt{3}}\sum_{n=1}^{\infty}\frac{3^{n}}{2n-1}.
$$
Using Desmos Both of the series diverges so $I_{1}$ is of the form $\infty -\infty$ which have a finite answer. Same thing goes with $I_{2}$.
Attempt:- 2:
Try to convert one integral into another. Substitute $x=\frac{1}{x}$ in $I_{1}$, we get
$$I_{1}=-\frac{3\pi}{2}\int_{1}^{\infty}\frac{x}{x^2+3} \,\mathrm dx
 +3\int_{1}^{\infty}\frac{x \space tan^{-1}x}{x^2+3} \,\mathrm dx $$
$$\implies I=-\frac{3\pi}{2}\int_{1}^{\infty}\frac{x}{x^2+3} \,\mathrm dx -\int_{0}^{\infty}\frac{x \space tan^{-1}x}{x^2+3} \,\mathrm dx  +4\int_{1}^{\infty}\frac{x \space tan^{-1}x}{x^2+3} \,\mathrm dx$$
Surprisingly all three integrals diverges and convergence of $I$ is maintained by negative and positive sign.
How can I prove the original result?
Thank you for your help!
 [/INST] Integrate by parts and then substitute $x=\tan \frac t2$
\begin{align}
I=& \>\int_{0}^{1}\frac{x\arctan x}{x^2+\frac13}\,\mathrm dx
-\int_{0}^{1}\frac{x\arctan x}{x^2+3}dx
=\frac14 \int_0^{\frac\pi2} \ln\frac{1+\frac12 \cos t}{1-\frac12 \cos t}dt
\end{align}
Let $J(a) = \int_0^{\frac\pi2}\ln(1+\cos a\cos t)dt$. Then
$$J’(a)
=-\tan a\left( \frac\pi2-\int_0^{\frac\pi2}\frac{1}{1+\cos a\cos t}dt\right) 
=a\sec a-\frac\pi2\tan a
$$
and
\begin{align}
I&= \frac14\left(J(\frac\pi3)-J(\frac{2\pi}3)\right)=-\frac14 \int^{\frac{\pi}2}_{\frac\pi3} J’(a)da - \frac14\int^{\frac{2\pi}3}_{\frac\pi2} \overset{a\to \pi -a}{J’(a)da }\\
 &=\frac12 \int_{\frac\pi3}^{\frac\pi2} (\frac\pi2-a)\sec a \>da 
\overset{a=\frac\pi2 - 2t}=2\int_0^{\frac\pi{12}} t\csc (2t) dt\\
&=- t\ln(\cot t)\bigg|_0^{\frac\pi{12}} + \int_{0} ^{\frac\pi{12}} 
{\ln(\cot t) dt}\\
 &= -\frac\pi{12} \ln(2+\sqrt3)+\frac23 G
\end{align}
where $\int^{ \frac\pi{12}}_{0} 
\ln(\cot t)dt= \frac23G$
</s>