Answer
$${\text{The integral is divergent}}$$
Work Step by Step
$$\eqalign{
& \int_2^\infty {\frac{1}{{x\ln x}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{}} \cr
& \int_2^\infty {\frac{1}{{x\ln x}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{1}{{x\ln x}}} dx \cr
& \cr
& {\text{integrating }}\int {\frac{1}{{x\ln x}}} dx \cr
& {\text{set }}u = \ln x{\text{ then }}\frac{{du}}{{dx}} = \frac{1}{x},\,\,\,\,xdu = dx \cr
& \int {\frac{1}{{x\ln x}}} dx = \int {\frac{1}{{xu}}} \left( {xdu} \right) = \int {\frac{1}{u}} du \cr
& = \ln \left| u \right| + C \cr
& {\text{replace }}u = \ln x \cr
& = \ln \left| {\ln x} \right| + C \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{1}{{x\ln x}}} dx = \mathop {\lim }\limits_{b \to \infty } \left( {\ln \left| {\ln x} \right|} \right)_2^b \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left( {\ln \left| {\ln b} \right| - \ln \left| {\ln 2} \right|} \right) \cr
& {\text{evaluating the limit when }}b \to \infty \cr
& = \ln \left| {\ln \infty } \right| - \ln \left| {\ln 2} \right| \cr
& = \infty \cr
& then \cr
& {\text{The integral is divergent}} \cr} $$
