Answer
$$\infty \,\,\,\left( {{\text{diverges}}} \right)$$
Work Step by Step
$$\eqalign{
& \int_2^{ + \infty } {\frac{1}{{x\sqrt {\ln x} }}} dx \cr
& \cr
& {\text{using the definition 7}}{\text{.8}}{\text{.1 of improper integrals}} \cr
& \,\,\,\int_a^{ + \infty } {f\left( x \right)} dx = \mathop {\lim }\limits_{b \to + \infty } \int_a^b {f\left( x \right)} dx \cr
& \cr
& {\text{then}} \cr
& \,\,\,\int_2^{ + \infty } {\frac{1}{{x\sqrt {\ln x} }}} dx = \mathop {\lim }\limits_{b \to + \infty } \int_2^b {\frac{1}{{x\sqrt {\ln x} }}} dx \cr
& = \mathop {\lim }\limits_{b \to + \infty } \int_2^b {\left( {\frac{1}{{\sqrt {\ln x} }}} \right)\left( {\frac{1}{x}} \right)} dx \cr
& = \mathop {\lim }\limits_{b \to + \infty } \int_2^b {{{\left( {\ln x} \right)}^{ - 1/2}}\left( {\frac{1}{x}} \right)} dx \cr
& \cr
& {\text{integrate by tables using the formula }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}}} + C \cr
& = \mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{{{{\left( {\ln x} \right)}^{1/2}}}}{{1/2}}} \right]_2^b \cr
& = 2\mathop {\lim }\limits_{b \to + \infty } \left[ {\sqrt {\ln x} } \right]_2^b \cr
& = 2\mathop {\lim }\limits_{b \to + \infty } \left[ {\sqrt {\ln b} - \sqrt {\ln 2} } \right] \cr
& \cr
& {\text{calculate the limit when }}b \to + \infty \cr
& = 2\left[ {\sqrt {\ln \infty } - \sqrt {\ln 2} } \right] \cr
& = 2\left( {\infty - \sqrt {\ln 2} } \right) \cr
& = \infty \,\,\,\left( {{\text{diverges}}} \right) \cr
& {\text{then}}{\text{,}} \cr
& {\text{In this case the integral diverges}} \cr} $$
