Answer
$1$
(converges)
Work Step by Step
$\displaystyle \int\frac{1}{x(\ln x)^{2}}dx=\left[\begin{array}{l}
u=\ln x\\
du=\frac{1}{x}dx
\end{array}\right]=\int u^{-2}du=\frac{u^{-1}}{-1}+C$
$=-\displaystyle \frac{1}{\ln x}+C$
The integral is a type I improper integral, so
$\displaystyle \int_{e}^{\infty}\frac{1}{x(\ln x)^{2}}dx=\lim_{t\rightarrow\infty}\int_{e}^{t}\frac{1}{x(\ln x)^{2}}dx$
$=\displaystyle \lim_{t\rightarrow\infty}\left[-\frac{1}{\ln x}\right]_{e}^{t}$
$=0-(-1)$
$=1$
(converges)
