Interesting integral 5 Find an integral:
$$
\int\limits_0^{+\infty} \frac{x^3 \ln^2 x}{1+x^7} d x.
$$
Should I use residue theorem? May somebody suggest any ideas to find it?
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\begin{equation}
\mbox{Note that}\quad
\int_{0}^{\infty}{x^{3}\ln^{2}\pars{x} \over 1 + x^{7}}\,\dd x =
\left.\partiald[2]{}{\mu}\int_{0}^{\infty}{x^{\mu} \over 1 + x^{7}}\,\dd x
\,\right\vert_{\ \mu\ =\ 3}
\label{1}\tag{1}
\end{equation}

With $\ds{x \equiv \pars{{1 \over t} - 1}^{1/7} \implies
t = {1 \over 1 + x^{7}}}$:
\begin{align}
\int_{0}^{\infty}{x^{\mu} \over 1 + x^{7}}\,\dd x & =
\int_{1}^{0}t\pars{{1 \over t} - 1}^{\mu/7}
\bracks{{1 \over 7}\pars{{1 \over t} - 1}^{-6/7}\pars{-\,{1 \over t^{2}}}}\dd t
\\[6mm] & =
{1 \over 7}\int_{0}^{1}t^{-\pars{\mu + 1}/7}\pars{1 - y}^{\pars{\mu - 6}/7}
\,\dd t =
{1 \over 7}\,{\Gamma\pars{\bracks{6 - \mu}/7}\Gamma\pars{\bracks{\mu + 1}/7} \over \Gamma\pars{1}}
\\[5mm] & =
{1 \over 7}\,{\pi \over \sin\pars{\pi\bracks{\mu + 1}/7}} =
{1 \over 7}\,\pi\csc\pars{{\pi \over 7}\,\bracks{\mu + 1}}
\end{align}


After a 'laborious' procedure; I'll find, by replacing in \eqref{1},

$$\bbx{\ds{%
\int_{0}^{\infty}{x^{3}\ln^{2}\pars{x} \over 1 + x^{7}}\,\dd x =
{\pi^{3} \over 686}\bracks{3 - \cos\pars{\pi \over 7}}
\sec^{3}\pars{\pi \over 14}}} \approx 0.1024
$$
