$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{-a}^{a}{x^{2} \over x^{4} + 1}\,\dd x:\ {\large ?}.\qquad a \in {\mathbb R}}$. $ \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}}}} $