Answer
Divergent
Work Step by Step
Define $f(x)=\frac{1}{x-\ln x}$ and $g(x)=\frac{1}{x}$.
For $x\geq 2$, we have
$\Leftrightarrow \ln x>0$
$\Leftrightarrow -\ln x< 0$
$\Leftrightarrow x-\ln x\frac{1}{x}$
$\Leftrightarrow f(x)>g(x)$
Find the convergence for $\int_2^\infty g(x) dx$:
$\int_2^\infty \frac{1}{x} dx=\lim\limits_{t \to \infty}\int_2^t\frac{1}{x} dx=\lim\limits_{t \to \infty}[\ln x]_2^t=\lim\limits_{t \to \infty}[\ln t-\ln 2]=\infty$
Since $f(x)>g(x)$ for $x\geq 2$ and $\int_2^\infty g(x)$ is divergent, using the Comparison Theorem it concludes that $\int_2^\infty \frac{1}{x-\ln x} dx$ is divergent.
