= Solution
Close the interval $[-R,R]$ with an upper <semicircle>, traversed counterclockwise. On its large arc,
$$
\left|\frac{z^2}{1+z^4}\right|\leq\frac{R^2}{R^4-1},
$$
so the <estimation lemma> bounds the arc integral by $\pi R^3/(R^4-1)\to0$. The enclosed <residues> are those at $(1+i)/\sqrt2$ and $(-1+i)/\sqrt2$, whose sum is $-i/(2\sqrt2)$. The <residue theorem> therefore gives
$$
\boxed{\int_{-\infty}^{\infty}\frac{x^2}{1+x^4}\,dx=2\pi i\left(-\frac{i}{2\sqrt2}\right)=\frac\pi{\sqrt2}.}
$$
The positive sign agrees with the nonnegative real integrand.
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