I am trying to evaluate this integral:$$K = \displaystyle \int\limits_0^1 {\frac{{{{\arctan }^2}\left( {\frac{x}{{1 - {x^2}}}} \right)}}{x}dx}$$
And here is my attempt so far:
$$\begin{align} K&=\int\limits_0^1 {\frac{{{{\arctan }^2}\left( {\frac{x}{{1 - {x^2}}}} \right)}}{x}dx} \xrightarrow{{{\text{IBP}}}} - 2\int\limits_0^1 {\frac{{\left( {{x^2} + 1} \right)\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx}\\&= - 2\int\limits_0^1 {\frac{{{x^2}\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} - 2\int\limits_0^1 {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} \hfill \\ \\&=- 2\int\limits_0^1 {\frac{{{x^2}\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} - 2\int\limits_0^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} \\&\quad\quad+ 2\underbrace {\int\limits_1^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} }_{x \to \frac{1}{x}} \\&= - 2\int\limits_0^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} \hfill \\ \end{align}$$
The last integral seems can be evaluated by using complex integration technique. But I don't know how to use it. And can I ask for help or another approach? Thank you so much.
The closed form for $K$ is: $$K=\frac{4\pi}{3}\cdot G-\frac{35}{36}\cdot\zeta(3)-\frac{5\pi}{36\sqrt{3}}(\psi'(1/3)-\psi'(2/3))$$
