Answer
Diverges.
Work Step by Step
Since
\begin{align*}
\int_{1}^{2} \frac{1}{x \ln x} d x&= \lim _{a \rightarrow 1^{+}}\int_{a}^{2} \frac{1}{x \ln x} d x\\
&=\ln [\ln x]\bigg|_a^2\\
&=\lim _{a \rightarrow 1^{+}}(\ln (\ln 2)-\ln (\ln a))\\
&=\infty
\end{align*}
Then the integral diverges.