= Solution
Close the real-line contour by a positively oriented upper semicircle of radius $R$, beyond every pole. The rational <function> is $O(R^{-2})$ uniformly on that arc, and $|e^{iz}|=e^{-\Im z}\leq1$. Its arc <integral> is therefore $O(R^{-1})$ and vanishes. The real <integral> is absolutely convergent, while no pole lies on it. The residue theorem gives
$$
\boxed{\int_{-\infty}^{\infty}F(x)e^{ix}\,dx
=2\pi i\sum_{\Im z_j>0}\operatorname{Res}_{z=z_j}\bigl(F(z)e^{iz}\bigr).}
$$
Only actual poles contribute; any removable zero of the denominator has zero residue. The same contour proof covers poles of any order.
For the particular denominator, the upper poles are $a=1+i$ and $b=-1+i$. They are simple, with residues $e^{ia}/(4a^3)$ and $e^{ib}/(4b^3)$. Since
$$
\frac1{4a^3}=\frac{-1-i}{16},\qquad
\frac1{4b^3}=\frac{1-i}{16},
$$
their sum is
$$
\frac{e^{-1}}{16}\bigl[(-1-i)e^{i}+(1-i)e^{-i}\bigr]
=-\frac{i e^{-1}}8(\cos1+\sin1).
$$
Multiplication by $2\pi i$ gives the real answer. The sine part of the real-line <integral> is zero by oddness, so
$$
\boxed{\int_{-\infty}^{\infty}\frac{\cos x}{4+x^4}\,dx
=\frac{\pi}{4e}(\cos1+\sin1).}
$$
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