Answer
The given integral
$$
\int_{2}^{\infty} \frac{dx}{x \ln x} =\infty,
$$
is divergent.
Work Step by Step
The given integral
$$
\int_{2}^{\infty} \frac{dx}{x \ln x}
$$
is an improper integral, hence
$$
\int_{2}^{\infty} \frac{d x}{x \ln x}=\lim _{t \rightarrow \infty} \int_{2}^{t} \frac{d x}{x \ln x}
$$
First, we evaluate the Indefinite Integral
$$
\int \frac{dx}{x \ln x}
$$
Let $ u = \ln x $. Then $ du = \frac{dx}{x}$ , so
$$
\int \frac{dx}{x \ln x} =\int \frac{d u}{u}=\ln |u|+C=\ln |\ln x|+C
$$
Now, we evaluate the improper integral:
$$
\begin{aligned}
\int_{2}^{\infty} \frac{d x}{x \ln x} &=\lim _{t \rightarrow \infty} \int_{2}^{t} \frac{d x}{x \ln x} \\
&=\lim _{t \rightarrow \infty}[\ln |\ln x|]_{2}^{t}\\
&=\lim _{t \rightarrow \infty}[\ln (\ln t)-\ln (\ln 2)] \\
&=\infty,
\end{aligned}
$$
so the given integral is divergent.