Answer
Divergent
Work Step by Step
Let \[I=\int_{0}^{1}\frac{1}{x}dx\]
Since 0 is point of infinite discontinuity of integrand $\Large\frac{1}{x}$
\[I=\int_{0}^{1}\frac{1}{x}dx=\lim_{t\rightarrow 0^+}\int_{t}^{1}\frac{1}{x}dx\]
\[I=\lim_{t\rightarrow 0^+}\int_{t}^{1}\frac{1}{x}dx\;\;\;\ldots(1)\]
\[I=\lim_{t\rightarrow 0^+}\left[\ln x\right]_{t}^{1}\]
\[I=\lim_{t\rightarrow 0^+}\left[0-\ln t\right]\]
$\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;$ does not exist
Since limit on R.H.S. of (1) does not exist so $I$ is divergent.
