Arctangent integral that I'm having difficulty on 
Question: Show that$$\int\limits_0^1 dx\,\frac {\arctan x}{\sqrt{x(1-x)}}=\pi\arctan\sqrt{\frac {\sqrt2-1}2}$$

I'm just having a hard time figuring out what to do. I tried to make the substitution $x=\frac {1-t}{1+t}$ but that didn't help very much because the denominator is slightly different. My next thought was to try to represent $\arctan x$ as an infinite series$$\arctan x=\sum\limits_{n\geq1}\frac {(-1)^{n-1}x^n}n\sin\left(\frac {\pi n}2\right)$$
But seeing as to how the result is in terms of $\arctan(\cdot)$, I doubt an infinite series would help much. Especially if the argument is a nested radical. Perhaps there is some sort of hidden symmetry one may exploit for this one?
 A: We have
$$ \int_{0}^{1}\frac{x^{2n+1}}{\sqrt{x(1-x)}}\,dx = \frac{\pi}{4^{2n}}\binom{4n}{2n}\frac{4n+1}{4n+2}$$
hence the given integral equals
$$ \frac{\pi}{2} \sum_{n\geq 0}\frac{(-1)^n}{4^{2n}}\binom{4n}{2n}\frac{4n+1}{(2n+1)^2} $$
where by the generating function for Catalan numbers we have
$$ \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\frac{2n+1}{n+1}z^n = \frac{2}{z\sqrt{1-z}}-\frac{2}{z}\tag{A}$$
hence by integration
$$ \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\frac{2n+1}{(n+1)^2}z^{n+1} = 4\log\left(\frac{2}{1+\sqrt{1-z}}\right)\tag{B}$$
and the given result can be proved by evaluating $(B)$ at $z=\pm i$, with some care in managing the determinations of the complex logarithm / square root. In a equivalent form
$$\int_{0}^{1}\frac{\arctan x}{\sqrt{x(1-x)}}\,dx = \pi\arctan\left(2^{1/4}\sin\tfrac{\pi}{8}\right).$$ 
A: For a straightforward approach, use that
$$\int_0^1 \frac{x\arccos(x)}{1+a x^2}\text{d}x=\frac{\pi}{2a}\log((1+\sqrt{1+a})/2),\tag{1}$$
since your integral is (by the integration by parts and a variable change) $\displaystyle 4\int_0^1 \frac{x\arccos(x)}{1+x^4}\text{d}x$. 
$\textbf{Q.E.D.}$
NOTE: the integral in $(1)$ is straightforward with the integration by parts and D.U.I.S. 
A: \begin{align}
\int_0^1\frac{\arctan x}{\sqrt{x(1-x)}} \,dx 
&= \int_0^1\frac{1}{\sqrt{x(1-x)}}\left(x-\dfrac{x^3}{3}+\dfrac{x^5}{5}+\cdots\right)\,dx\\
&= \int_0^1 \left(x^\frac{1}{2}(1-x)^\frac{-1}{2}-\dfrac13x^\frac{5}{2}(1-x)^\frac{-1}{2} + \dfrac15x^\frac{9}{2}(1-x)^\frac{-1}{2}-\cdots\right)\,dx\\
&= \beta\left(\dfrac{3}{2},\dfrac{1}{2}\right) -\frac13\beta\left(\dfrac{7}{2},\dfrac{1}{2}\right)+\frac15\beta\left(\dfrac{11}{2},\dfrac{1}{2}\right)-\cdots\\
&= \pi\sum_{n=0}^{\infty}\dfrac{(-1)^n}{(2n+1)2^{4n+1}}{4n+1\choose2n+1}
\end{align}
