Answer
$${\text{The improper integral diverges}}{\text{.}}$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{1}{{x\ln x}}} dx \cr
& \frac{1}{{x\ln x}}{\text{ has an infinite discontinuity at }}x = 1,{\text{so we can write}} \cr
& \int_1^\infty {\frac{1}{{x\ln x}}} dx = \mathop {\lim }\limits_{a \to {1^ + }} \int_a^2 {\frac{1}{{x\ln x}}} dx + \mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{1}{{x\ln x}}} dx \cr
& {\text{Integrating}} \cr
& = \mathop {\lim }\limits_{a \to {1^ + }} \left[ {\ln \left| {\ln x} \right|} \right]_a^2 + \mathop {\lim }\limits_{b \to \infty } \left[ {\ln \left| {\ln x} \right|} \right]_2^b \cr
& = \mathop {\lim }\limits_{a \to {1^ + }} \left[ {\ln \left| {\ln 2} \right| - \ln \left| {\ln a} \right|} \right] + \mathop {\lim }\limits_{b \to \infty } \left[ {\ln \left| {\ln b} \right| - \ln \left| {\ln 2} \right|} \right] \cr
& {\text{Evaluate the limits}} \cr
& = \ln \left| {\ln 2} \right| - \ln \left| {\ln {1^ + }} \right| + \ln \left| {\ln \infty } \right| - \ln \left| {\ln 2} \right| \cr
& = - \ln \left| {\ln {1^ + }} \right| + \ln \left| {\ln \infty } \right| \cr
& = - \ln \left| {{0^ + }} \right| + \ln \left| \infty \right| \cr
& {\text{The improper integral diverges}}{\text{.}} \cr} $$
