Proving $\int_{0}^{1} \frac{\tanh^{-1}\sqrt{x(1-x)}}{\sqrt{x(1-x)}}dx=\frac{1}{3}(8C-\pi\ln(2+\sqrt{3}))$ for an identity of Srinivasa Ramanujan Ramanujan is supposed to  have given more than five thousand elegant results, a good number of them are yet to be proved or disproved.
Yesterday in the comment section of
Proving that $ \sum_{k=0}^\infty\frac1{2k+1}{2k \choose k}^{-1}=\frac {2\pi}{3\sqrt{3}} $
A wonderful Ramanujan identity
$$S=\sum_{k=0}^{\infty}  \frac{1}{(2k+1)^2}{2k \choose k}^{-1}=\frac{1}{3}(8C-\pi\ln(2+\sqrt{3}))~~~~(1)$$ was showcased, Mathematica also gives this out.
My Attempt to prove (1) by hand:
Note the integral representation of the reciprocal of the binomial co-efficient:
$${n \choose j}^{-1}=(n+1)\int_{0}^{1} x^j (1-x)^{n-j}~ dx~~~~(2)$$
$$S=\sum_{k=0}^{\infty}  \frac{1}{(2k+1)^2}{2k \choose k}^{-1}= \int_{0}^{1} \sum_{k=0}^{\infty} \frac{[x(1-x)]^{k}}{(2k+1)} dx= \int_{0}^{1} \frac{\tanh^{-1}\sqrt{x(1-x)}}{\sqrt{x(1-x)}} dx~~~~(3)$$
The question is:  How to get this integral (3) by hand ?

You can give Feynman's trick a shot.
\begin{align*}
I&=\int _0^1\frac{\operatorname{arctanh} \left(\sqrt{x\left(1-x\right)}\right)}{\sqrt{x\left(1-x\right)}}\:dx\\[3mm]
I\left(a\right)&=\int _0^1\frac{\operatorname{arctanh} \left(a\sqrt{x\left(1-x\right)}\right)}{\sqrt{x\left(1-x\right)}}\:dx\\[3mm]
I'\left(a\right)&=\int _0^1\frac{1}{1-a^2x\left(1-x\right)}\:dx=\frac{4}{a\sqrt{4-a^2}}\arctan \left(\frac{a}{\sqrt{4-a^2}}\right)\\[3mm]
\int _0^1I'\left(a\right)da&=4\underbrace{\int _0^1\frac{1}{a\sqrt{4-a^2}}\arctan \left(\frac{a}{\sqrt{4-a^2}}\right)\:da}_{t=\frac{a}{\sqrt{4-a^2}}}\\[3mm]
I&=8\underbrace{\int _0^{\frac{1}{\sqrt{3}}}\frac{\arctan \left(t\right)}{4t\sqrt{1+t^2}}\:dt}_{t=\tan\left(x\right)}=2\int _0^{\frac{\pi }{6}}\frac{x\sec \left(x\right)}{\tan \left(x\right)}\:dx\\[3mm]
&=2\int _0^{\frac{\pi }{6}}\frac{x}{\sin \left(x\right)}\:dx
\end{align*}
That integral has been evaluated here by Zacky, using its result yields
$$\boxed{I=\int _0^1\frac{\operatorname{arctanh} \left(\sqrt{x\left(1-x\right)}\right)}{\sqrt{x\left(1-x\right)}}\:dx=\frac{\pi}{3}\ln(2-\sqrt 3) +\frac{8}{3}G}$$

One can also find the last integral by using the Weierstrass substitution.
\begin{align*}
2\int _0^{\frac{\pi }{6}}\frac{x}{\sin \left(x\right)}\:dx&=4\underbrace{\int _0^{2-\sqrt{3}}\frac{\arctan \left(t\right)}{t}\:\:dt}_{\operatorname{IBP}}\\[3mm]
&=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)-4\underbrace{\int _0^{2-\sqrt{3}}\frac{\ln \left(t\right)}{1+t^2}\:dt}_{t=\tan\left(x\right)}\\[2mm]
&=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)-4\int _0^{\frac{\pi }{12}}\ln \left(\tan \left(x\right)\right)\:dx\\[3mm]
&=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)+8\sum _{k=1}^{\infty }\frac{1}{2k-1}\int _0^{\frac{\pi }{12}}\cos \left(2\left(2k-1\right)x\right)\:dx\\[3mm]
&=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)+4\sum _{k=1}^{\infty }\frac{\sin \left(\frac{\pi }{6}\left(2k-1\right)\right)}{\left(2k-1\right)^2}\\[3mm]
&=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)+\frac{8}{3}\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{\left(2k-1\right)^2}\\[3mm]
&=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)+\frac{8}{3}G
\end{align*}
