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 ?

Note
$\int_{0}^{1} \frac{\tanh^{-1}\sqrt{x(1-x)}}{\sqrt{x(1-x)}}dx
\overset{x=\sin^2t} =\int_0^{\pi/2} 2
\tanh^{-1}\frac{\sin 2t}2dt
= \int_0^{\pi/2}\ln\left(\frac{1+\frac{\sin2t}2}{1-\frac{\sin2t}2}\right)dt
$
Solving the integral $\int_0^{\pi/2}\log\left(\frac{2+\sin2x}{2-\sin2x}\right)\mathrm dx$ $=\frac{1}{3}[8C-\pi\ln(2+\sqrt{3})]$
