Answer
Divergent.
Work Step by Step
The integrand has a discontinuity at 0, so this is a type 2.b improper integral.
$\displaystyle \int_{0}^{1}\frac{1}{x}dx=\lim_{t\rightarrow 0^{+}}\int_{t}^{1}\frac{1}{x}dx$
$=\displaystyle \lim_{t\rightarrow 0^{+}}[\ln|x|]_{t}^{1}$
$=\displaystyle \lim_{t\rightarrow 0^{+}}(-\ln t-0)$
$=+\infty.\quad $
