Formulae for Catalan's constant. Some years ago, someone had shown me the formula (1).
I have searched for its origin and for a proof.
I wasn't able to get true origin of this formula but I was able to find out an elementary proof for it.
Since then, I'm interested in different approaches to find more formulae as (1).
What other formulas similar to ($1$) are known?
Two days ago, reading the book of Lewin "Polylogarithms and Associated Functions" I was able to find out formula (2).
$\displaystyle \dfrac{1}{3}C=\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{x(1-x)}{2-x}\right)dx\tag1$ 
$\displaystyle \dfrac{2}{5}C=\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{\sqrt{5}x(1-x)}{1+\sqrt{5}-\sqrt{5}x}\right)dx-\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{x(1-x)}{3+\sqrt{5}-x}\right)dx\tag2$ 
$C$ being the Catalan's constant.
I have a proof for both of these formulae.
My approach relies on the following identity:
For all real $x>1$,
$\displaystyle \int_0^1 \dfrac{1}{t} \arctan \left (\dfrac{t(1-t)}{\frac{x+1}{2}-t}\right) dt=\int_1^{\frac{\sqrt{x}+1}{\sqrt{x}-1}}\dfrac{\log(t)}{1+t^2}dt$

A more natural proof.
\begin{align}
\beta&=\sqrt{3}-1\\    
J&=\int_0^1 \frac{\arctan\left(\frac{x(1-x)}{2-x}\right)}{x}dx\\
&\overset{\text{IBP}}=\left[\arctan\left(\frac{x(1-x)}{2-x}\right)\ln x\right]_0^1-\int_0^1 \frac{(x^2-4x+2)\ln x}{(x^2+\beta x+2)(x^2-(\beta+2)x+2)}dx\\
&=-\int_0^1 \frac{(x^2-4x+2)\ln x}{(x^2+\beta x+2)(x^2-(\beta+2)x+2)}dx\\
&=\int_0^1 \frac{\beta\ln x}{2(x^2-(2+\beta) x+2)}-\int_0^1 \frac{(2+\beta)\ln x}{2(x^2+\beta x+2)}\\
&=\underbrace{\int_0^1 \frac{2\ln x}{\beta\left(\left(\frac{2x-2-\beta}{\beta}\right)^2+1\right)}dx}_{y=\frac{\beta}{2+\beta-2x}}-\underbrace{\int_0^1 \frac{2\ln x}{(2+\beta)\left(\left(\frac{2x+\beta}{2+\beta}\right)^2+1\right)}dx}_{y=\frac{2x+\beta}{2+\beta}}\\
&=-\int_{\frac{\beta}{\beta+2}}^1 \frac{\ln y}{1+y^2}dy\\
&\overset{y=\tan \theta}=-\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\ln\left(\tan \theta\right)d\theta\\
&=-\int_{0}^{\frac{\pi}{4}}\ln\left(\tan \theta\right)d\theta+\int_0^{\frac{\pi}{12}}\ln\left(\tan \theta\right)d\theta\\
&=\text{G}-\frac{2}{3}\text{G}\\
&=\boxed{\dfrac{1}{3}\text{G}}
\end{align}
NB: For the latter integral see Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$
