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