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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

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

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

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    $\begingroup$ You might enjoy reading this paper on inverse trig integrals by Rogers: arxiv.org/abs/math/0601082 . Your integral is equivalent to a special case of the third family of integrals evaluated there. $\endgroup$
    – David H
    Feb 18, 2016 at 13:26
  • $\begingroup$ @DavidH thank you. Indeed, nice paper (+1). $\endgroup$
    – user 1591719
    Feb 18, 2016 at 13:47

1 Answer 1

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

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  • $\begingroup$ Lovely stuff +1 $\endgroup$
    – user284001
    Feb 18, 2016 at 13:12

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