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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

$$\int_0^{1} \frac{\displaystyle \arctan^2\left(\frac{\sqrt{x}}{\sqrt{4-x}}\right)}{x} \, dx$$

$$=\frac{1}{216} \left(\sqrt{3} \pi \left(\psi ^{(1)}\left(\frac{1}{3}\right)-\psi ^{(1)}\left(\frac{2}{3}\right)+\psi ^{(1)}\left(\frac{1}{6}\right)-\psi ^{(1)}\left(\frac{5}{6}\right)\right)-144 \zeta (3)\right).$$

user 1591719
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    You might enjoy reading this paper on inverse trig integrals by Rogers: http://arxiv.org/abs/math/0601082 . Your integral is equivalent to a special case of the third family of integrals evaluated there. – David H Feb 18 '16 at 13:26
  • @DavidH thank you. Indeed, nice paper (+1). – user 1591719 Feb 18 '16 at 13:47
  • $$\int_0^{1} \frac{\arctan^2\frac{\sqrt{x}}{\sqrt{4-x}}}{x} , dx=\int_0^{\frac\pi6}2t^2 \cot t dt =-\frac23\zeta(3)-\frac\pi3 \Im Li_2 (\frac{1-i\sqrt3}2) $$ – Quanto Jan 29 '23 at 14:32

1 Answers1

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This is not even an idea, it's an evidence: set $$x=4\sin^2\phi.$$

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