TITLE: How to prove $\int_0^1\frac{x^3\arctan x}{(3-x^2)^2}\frac{\mathrm dx}{\sqrt{1-x^2}}=\frac{\pi\sqrt{2}}{192}\left(18-\pi-6\sqrt{3}\,\right)$?
QUESTION [18 upvotes]: How to prove the following result?

$$\int_0^1\frac{x^3\arctan x}{(3-x^2)^2}\frac{\mathrm dx}{\sqrt{1-x^2}}=\frac{\pi\sqrt{2}}{192}\left(18-\pi-6\sqrt{3}\,\right)$$

For my part no idea?

REPLY [3 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{0}^{1}\frac{x^{3}\arctan\pars{x}}{\pars{3 - x^{2}}^{2}}
     \frac{\dd x}{\root{1 - x^{2}}}=\frac{\pi\root{2}}{192}\,
     \pars{18 - \pi - 6\root{3}} \approx {\tt 0.1033}:\ {\large ?}}$

\begin{align}&\overbrace{\color{#66f}{\large%
\int_{0}^{1}\frac{x^{3}\arctan\pars{x}}{\pars{3 - x^{2}}^2}
\frac{\dd x}{\root{1 - x^{2}}}}}^{\ds{\dsc{x}\ \mapsto\ \dsc{\frac{1}{x}}}}\ =\
\int_{1}^{\infty}\frac{\arctan\pars{1/x}}{\pars{3x^{2} - 1}^{2}}
\frac{\dd x}{\root{x^{2} - 1}}
\\[5mm]&=\int_{1}^{\infty}\ \overbrace{\int_{0}^{1}\frac{x\,\dd t}{t^{2} + x^{2}}}
^{\dsc{\arctan\pars{1/x}}}\
\frac{\dd x}{\pars{3x^{2} - 1}^{2}\root{x^{2} - 1}}
=\int_{0}^{1}\ \overbrace{\int_{1}^{\infty}
\frac{x\,\dd x}{\pars{x^{2} + t^{2}}\pars{3x^{2} - 1}^{2}\root{x^{2} - 1}}}
^{\ds{\dsc{x}\ \mapsto\ \dsc{\root{x}}}}\ \,\dd t
\\[5mm]&=\half\int_{0}^{1}\int_{1}^{\infty}
\frac{\dd x}{\pars{x + t^{2}}\pars{3x - 1}^{2}\root{x - 1}}\,\dd t
\\[5mm]&=\left.\frac{1}{18}\,\totald{}{\mu}\int_{0}^{1}\int_{1}^{\infty}
\frac{\dd x}{\pars{x + t^{2}}\pars{x - \mu}\root{x - 1}}\,\dd t
\right\vert_{\mu\ =\ 1/3}
\\[5mm]&=\left.\frac{1}{18}\,\totald{}{\mu}\int_{0}^{1}
\frac{1}{t^{2} + \mu}\braces{%
\underbrace{\int_{1}^{\infty}\frac{\dd x}{\pars{x - \mu}\root{x - 1}}}
_{\dsc{\frac{\pi}{\root{1 - \mu}}}}\ -\
\underbrace{\int_{1}^{\infty}\frac{\dd x}{\pars{x + t^{2}}\root{x - 1}}}
_{\dsc{\frac{\pi}{\root{t^{2} + 1}}}}}\,\dd t
\right\vert_{\mu\ =\ 1/3}
\end{align}
The '$\ds{x}$-integrations' were easily evaluated with the change
$\ds{\dsc{\root{x - 1}}\ \mapsto\ \dsc{x}}$. Note that
$\ds{\int_{0}^{1}\frac{\dd t}{t^{2} + \mu}
     =\frac{\arctan\pars{\mu^{-1/2}}}{\root{\mu}}}$. Namely,
\begin{align}&\color{#66f}{\large%
\int_{0}^{1}\frac{x^{3}\arctan\pars{x}}{\pars{3 - x^{2}}^2}
\frac{\dd x}{\root{1 - x^{2}}}}
\\[5mm]&=\overbrace{\frac{\pi}{18}\,\totald{}{\mu}\bracks{%
\frac{\arctan\pars{\mu^{-1/2}}}{\root{\mu\pars{1 - \mu}}}}_{\mu\ =\ 1/3}}
^{\dsc{-\,\frac{\pi\root{2}}{192}\,\pars{4\pi + 6\root{3}}}}\ -\,\,\,
\overbrace{\left.\frac{\pi}{18}\,\totald{}{\mu}
\int_{0}^{1}\frac{\dd t}{\pars{t^{2} + \mu}\root{t^{2} + 1}}
\right\vert_{\mu\ =\ 1/3}}
^{\dsc{-\,\frac{\pi\root{2}}{192}\,\pars{3\pi + 18}}}
\\[5mm]&=\color{#66f}{\large\frac{\pi\root{2}}{192}\,
\pars{18 - \pi - 6\root{3}}} \approx {\tt 0.1033}
\end{align}
The last integral is evaluated as follows:
\begin{align}&\overbrace{%
\int_{0}^{1}\frac{\dd t}{\pars{t^{2} + \mu}\root{t^{2} + 1}}}
^{\ds{\dsc{t}\ \mapsto\ \dsc{\frac{1}{t}}}}
=\int_{1}^{\infty}\frac{t\,\dd t}{\pars{1 + \mu t^{2}}\root{t^{2} + 1}}
=\half\int_{1}^{\infty}\frac{\dd t}{\pars{1 + \mu t}\root{t + 1}}
\\[5mm]&=\frac{1}{2\mu}
\int_{2}^{\infty}\frac{\dd t}{\pars{t - 1 + 1/\mu}\root{t}}
=\frac{1}{\mu}
\int_{\root{2}}^{\infty}\frac{\dd t}{t^{2} + 1/\mu - 1}
=\frac{\arctan\pars{\sqrt{\frac{1}{2\mu}-\half}}}
{\sqrt{\mu\pars{1 - \mu}}}
\end{align}