Arctangent integral that I'm having difficulty on

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 $$\int_{0}^{1}\frac{\arctan x}{\sqrt{x(1-x)}}\,dx = \pi\arctan\left(2^{1/4}\sin\tfrac{\pi}{8}\right).$$

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.

\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}