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\begin{align}
{\cal F}\pars{\mu}&\equiv
\int_0^{\infty}{\dd x \over \mu + x^{2}}
=\mu^{-1/2}\int_0^{\infty}{\dd x \over 1 + x^{2}} = \half\,\pi\mu^{-1/2}
\end{align}
\begin{align}
{\cal F}^{'''}\pars{\mu}&\equiv
-3!\int_0^{\infty}{\dd x \over \pars{\mu + x^{2}}^{4}}
=\half\,\pi\,\totald[3]{\mu^{-1/2}}{\mu}
=\half\,\pi\,\pars{-\,\half}\pars{-\,{3 \over 2}}\pars{-\,{5 \over 2}}\mu^{-7/2}
\end{align}
Set $\mu = 1$ in both members:
$$
\color{#00f}{\large\int_0^{\infty}{\dd x \over \pars{1 + x^{2}}^{4}}}
={15\pi/16 \over 6}= \color{#00f}{\large{5 \over 32}\,\pi}
$$