What is your best idea for this integral? No need for any full solution, this is optional. I'm curious
about the core idea you might like to use to make all very simple.
Calculate in closed-form
$$\int_0^{1} \frac{\displaystyle \arctan^2\left(\frac{\sqrt{x}}{\sqrt{4-x}}\right)}{x} \, dx$$
Supplementary questions:
$$a) \ \int_0^{1} \frac{\displaystyle \arctan^3\left(\frac{\sqrt{x}}{\sqrt{4-x}}\right)}{x} \, dx$$
$$b) \ \int_0^{1} \frac{\displaystyle \arctan^3\left(\frac{\sqrt{x}}{\sqrt{4-x}}\right)\log(x)}{x} \, dx.$$
EDIT: Maybe it helps to write down the closed-form I got for the first integral, which is
$$\int_0^{1} \frac{\displaystyle \arctan^2\left(\frac{\sqrt{x}}{\sqrt{4-x}}\right)}{x} \, dx$$
$$=\frac{1}{216} \left(\sqrt{3} \pi \left(\psi ^{(1)}\left(\frac{1}{3}\right)-\psi ^{(1)}\left(\frac{2}{3}\right)+\psi ^{(1)}\left(\frac{1}{6}\right)-\psi ^{(1)}\left(\frac{5}{6}\right)\right)-144 \zeta (3)\right).$$