= Solution
On the upper semicircular <contour>, $|z|=R$ and the <reverse triangle inequality> gives $|z^4+1|\ge R^4-1>0$. Its <arc length> is $\pi R$, so the <estimation lemma> gives
$$
\boxed{\left|\int_{\gamma_R}\frac{z^2}{z^4+1}\,dz\right|\le\frac{\pi R^3}{R^4-1}\longrightarrow0.}
$$
The arc is open, so an ordinary integral sign is appropriate even though the source uses a closed-integral symbol there. Subtracting the vanishing arc contribution from the closed <contour integral> gives the real-line <improper integral> $\pi/\sqrt2$. Its integrand is an <even function>, and its tail is $O(x^{-2})$, ensuring convergence. Thus
$$
\boxed{\int_0^\infty\frac{x^2}{x^4+1}\,dx=\frac{\pi}{2\sqrt2}.}
$$
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