Answer
$$\frac{1}{2}$$
Work Step by Step
$$\eqalign{
& \int_2^\infty {\frac{1}{{x{{\left( {\ln x} \right)}^2}}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{}} \cr
& \int_2^\infty {\frac{1}{{x{{\left( {\ln x} \right)}^2}}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{1}{{x{{\left( {\ln x} \right)}^2}}}} dx \cr
& \cr
& {\text{integrating }}\int {\frac{1}{{x{{\left( {\ln x} \right)}^2}}}} dx \cr
& {\text{set }}u = \ln x{\text{ then }}\frac{{du}}{{dx}} = \frac{1}{x},\,\,\,\,xdu = dx \cr
& \int {\frac{1}{{x{{\left( {\ln x} \right)}^2}}}} dx = \int {\frac{1}{{x{u^2}}}} \left( {xdu} \right) = \int {\frac{1}{{{u^2}}}} du \cr
& = - \frac{1}{u} + C \cr
& {\text{replace }}u = \ln x \cr
& = - \frac{1}{{\ln x}} + C \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{1}{{x{{\left( {\ln x} \right)}^2}}}} dx = \mathop {\lim }\limits_{b \to \infty } \left( { - \frac{1}{{\ln x}}} \right)_2^b \cr
& = - \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{{\ln x}}} \right)_2^b \cr
& {\text{evaluating the limit when }}b \to \infty \cr
& = - \left( {\frac{1}{{\ln \infty }} - \frac{1}{2}} \right) \cr
& = - \left( {0 - \frac{1}{2}} \right) \cr
& = \frac{1}{2} \cr} $$
