Solution
= Solution
Parametrize the positively oriented <unit circle> by $z=e^{i\theta}$. Then $dz=iz\,d\theta$ and
$$
\frac{z^4-10z^2+1}{z^2}=2\cos(2\theta)-10=-4(2+\sin^2\theta).
$$
Consequently the <contour integral> already evaluated satisfies
$$
I=-\frac i4\int_0^{2\pi}\frac{d\theta}{2+\sin^2\theta},\qquad
\boxed{\int_0^{2\pi}\frac{d\theta}{2+\sin^2\theta}=4iI=\frac{2\pi}{\sqrt6}.}
$$