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$\ds{\int_{-a}^{a}{x^{2} \over x^{4} + 1}\,\dd x:\ {\large ?}.\qquad
a \in {\mathbb R}}$.
$$
\color{#c00000}{\int_{-a}^{a}{x^{2} \over x^{4} + 1}\,\dd x}
=2\sgn\pars{a}\int_{0}^{\verts{a}}{x^{2} \over x^{4} + 1}\,\dd x
$$
With
$\ds{t \equiv {1 \over x^{4} + 1}\quad\imp\quad x = \pars{{1 \over t} - 1}^{1/4}}$:
\begin{align}
&\color{#c00000}{\int_{-a}^{a}{x^{2} \over x^{4} + 1}\,\dd x}
=
2\sgn\pars{a}\int_{1}^{1/\pars{a^{4} + 1}}t\pars{{1 \over t} - 1}^{1/2}
\bracks{{1 \over 4}\,\pars{{1 \over t} - 1}^{-3/4}\,\pars{-\,{1 \over t^{2}}}\,\dd t}
\\[3mm]&=\half\,\sgn\pars{a}\int^{1}_{1/\pars{a^{4} + 1}}
t^{-3/4}\pars{1 - t}^{-1/4}\,\dd t
\\[3mm]&=\half\,\sgn\pars{a}\bracks{%
\int^{1}_{0}t^{-3/4}\pars{1 - t}^{-1/4}\,\dd t
-
\int_{0}^{1/\pars{a^{4} + 1}}t^{-3/4}\pars{1 - t}^{-1/4}\,\dd t}
\\[3mm]&=\half\,\sgn\pars{a}\bracks{%
{\rm B}\pars{{1 \over 4},{3 \over 4}}
-{\rm B}\pars{{1 \over a^{4} + 1};{1 \over 4},{3 \over 4}}}
\end{align}
where $\ds{{\rm B}}$'s are Beta Functions.
Moreover,
$\ds{{\rm B}\pars{{1 \over 4},{3 \over 4}} = \Gamma\pars{1 \over 4}
\Gamma\pars{3 \over 4} = {\pi \over \sin\pars{\pi/4}} = \root{2}\,\pi}$.
$\ds{\Gamma\pars{z}}$ is the
Gamma Function and we used well known properties of $\ds{\rm B}$'s and $\ds{\Gamma}$'s.
$$
\color{#00f}{\large\int_{-a}^{a}{x^{2} \over x^{4} + 1}\,\dd x
=\half\,\sgn\pars{a}\bracks{\root{2}\pi
-{\rm B}\pars{{1 \over a^{4} + 1};{1 \over 4},{3 \over 4}}}}
$$