Answer
Diverges
Work Step by Step
$$\eqalign{
& \int_2^\infty {\frac{x}{{\sqrt {{x^2} - 1} }}} dx \cr
& {\text{Using the definition of improper integrals }} \cr
& \underbrace {\int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{t \to \infty } } \int_a^t {f\left( x \right)} dx}_ \Downarrow \cr
& \int_2^\infty {\frac{x}{{\sqrt {{x^2} - 1} }}} dx = \mathop {\lim }\limits_{t \to \infty } \int_2^t {\frac{x}{{\sqrt {{x^2} - 1} }}} dx \cr
& {\text{Rewrite the integrand}} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{t \to \infty } \int_2^t {\frac{{2x}}{{\sqrt {{x^2} - 1} }}} dx \cr
& = \frac{1}{2}\mathop {\lim }\limits_{t \to \infty } \int_2^t {{{\left( {{x^2} - 1} \right)}^{ - 1/2}}\left( {2x} \right)} dx \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{t \to \infty } \left[ {\frac{{{{\left( {{x^2} - 1} \right)}^{1/2}}}}{{1/2}}} \right]_2^t \cr
& = \mathop {\lim }\limits_{t \to \infty } \left[ {\sqrt {{x^2} - 1} } \right]_2^t \cr
& = \mathop {\lim }\limits_{t \to \infty } \left[ {\sqrt {{t^2} - 1} - \sqrt {{2^2} - 1} } \right] \cr
& = \mathop {\lim }\limits_{t \to \infty } \left[ {\sqrt {{t^2} - 1} - \sqrt 3 } \right] \cr
& {\text{Evaluate the limit when }}t \to \infty \cr
& = \infty - \sqrt 3 \cr
& = \infty \cr
& {\text{Therefore}}{\text{, the integral diverges.}} \cr} $$
