TITLE: 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
QUESTION [9 upvotes]: 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 ?

REPLY [3 votes]: If we denote the sum by $S$, we have the short proof

$$S=\sum_{k=0}^{\infty}  \int_0^1 \frac{x^{2k}}{\displaystyle (2k+1){2k \choose k}}\textrm{d}x=4\int_0^1\frac{\arcsin(x/2)}{x\sqrt{4-x^2}}\textrm{d}x=2\int_0^{\pi/6}\frac{x}{\sin(x)}\textrm{d}x$$
$$=4\int_0^{2-\sqrt{3}}\frac{\arctan(x)}{x}\textrm{d}x=4\operatorname{Ti}_2(2-\sqrt{3})=\frac{8}{3}G+\frac{\pi}{3}\log(2-\sqrt{3}).$$
Q.E.D.

$\operatorname{Ti}_2(2-\sqrt{3})$ is a special value of the inverse tangent integral that is extracted immediately by (also) using a famous result by Ramanujan,
$$\sum_{n=1}^{\infty} \frac{\sin(2(2n-1)x)}{(2n-1)^2}=\operatorname{Ti}_2(\tan(x))-x \log(\tan(x)), \ 0<x<\frac{\pi}{2},$$
and these details may be found in the book, (Almost) Impossible Integrals, Sums, and Series, pages $215$-$216$.
A note: To have a clear picture of the arcsine series used one can express $\displaystyle {2k \choose k}$ in terms of $\displaystyle {2k+2 \choose k+1}$ and then reindex the series. That's all.