= Solution
Assume positive, counterclockwise <orientation of a contour>. Factor the <polynomial> denominator as
$$
z^4-10z^2+1=\bigl(z^2-(5-2\sqrt6)\bigr)\bigl(z^2-(5+2\sqrt6)\bigr).
$$
Only $z=\pm(\sqrt3-\sqrt2)$ lie inside the <unit circle>. These are <simple poles>. Writing the denominator as $P(z)$, their <residues> are
$$
\frac{z}{P'(z)}=\frac1{4z^2-20}=-\frac1{8\sqrt6}.
$$
The <residue theorem> therefore gives
$$
\boxed{I=2\pi i\left(-\frac1{4\sqrt6}\right)=-\frac{\pi i}{2\sqrt6}.}
$$
Reversing <orientation of a contour> changes the sign.
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