Answer
Diverges
Work Step by Step
Use the direct limit comparison test.
Since, $ \dfrac{1}{ \sqrt {x^2-1}} \geq \dfrac{1}{x}$
Thus, $0\leq \dfrac{1}{x}\leq \dfrac{1}{ \sqrt {x^2-1}}$; as $x \to\infty$
Now, we have
$\int_{1}^{\infty} \dfrac{dx}{x}=\lim\limits_{a \to \infty} \int_{1}^{a} \dfrac{dx}{x} \\ \lim\limits_{a \to \infty} \int_{1}^{a} \dfrac{dx}{x} \\=\lim\limits_{a \to \infty} [\ln |x|]_{1}^{a} \\ \\=\lim\limits_{a \to \infty} [0-\ln |a|] \\=\infty$
Thus, the integral diverges by the direct limit comparison test.