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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:

$$\begin{gathered} K =\int\limits_0^1 {\frac{{{{\arctan }^2}\left( {\frac{x}{{1 - {x^2}}}} \right)}}{x}dx} \xrightarrow{{{\text{IBP}}}} - 2\int\limits_0^1 {\frac{{\left( {{x^2} + 1} \right)\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} = - 2\int\limits_0^1 {\frac{{{x^2}\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} - 2\int\limits_0^1 {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} \hfill \\ =- 2\int\limits_0^1 {\frac{{{x^2}\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} - 2\int\limits_0^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} + 2\underbrace {\int\limits_1^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} }_{x \to \frac{1}{x}} = - 2\int\limits_0^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} \hfill \\ \end{gathered}$$

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.

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: $$K=\frac{4\pi}{3}\cdot G-\frac{35}{36}\cdot\zeta(3)-\frac{5\pi}{36\sqrt{3}}(\psi'(1/3)-\psi'(2/3))$$

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:

$$\begin{array}{l} K = \displaystyle\int\limits_0^1 {\frac{{{{\arctan }^2}\left( {\frac{x}{{1 - {x^2}}}} \right)}}{x}dx} - 2\int\limits_0^1 {\frac{{\left( {{x^2} + 1} \right)\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} = - 2\int\limits_0^1 {\frac{{{x^2}\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} - 2\int\limits_0^1 {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} \\ = - 2\displaystyle\int\limits_0^1 {\frac{{{x^2}\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} - 2\int\limits_0^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} + 2\underbrace {\int\limits_1^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} }_{x \to \frac{1}{x}} = - 2\int\limits_0^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} \end{array}$$

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.

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:

$$\begin{gathered} K =\int\limits_0^1 {\frac{{{{\arctan }^2}\left( {\frac{x}{{1 - {x^2}}}} \right)}}{x}dx} \xrightarrow{{{\text{IBP}}}} - 2\int\limits_0^1 {\frac{{\left( {{x^2} + 1} \right)\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} = - 2\int\limits_0^1 {\frac{{{x^2}\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} - 2\int\limits_0^1 {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} \hfill \\ =- 2\int\limits_0^1 {\frac{{{x^2}\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} - 2\int\limits_0^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} + 2\underbrace {\int\limits_1^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} }_{x \to \frac{1}{x}} = - 2\int\limits_0^\infty {\frac{{\arctan \left( {\frac{x}{{1 - {x^2}}}} \right)\ln \left( x \right)}}{{{x^4} - {x^2} + 1}}dx} \hfill \\ \end{gathered}$$

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.