Answer
Divergent
Work Step by Step
Let \[I=\int_{0}^{1}\frac{1}{2-3x}dx\]
Clearly $\frac{2}{3}$ is point of infinite discontinuity of integrand $\large\frac{1}{2-3x}$
\[\Rightarrow I=\int_{0}^{\frac{2}{3}}\frac{1}{2-3x}dx+\int_{\frac{2}{3}}^{1}\frac{1}{2-3x}dx\;\;\;\ldots (1)\]
Let \[I_1=\lim_{t\rightarrow \frac{2}{3}^-}\int_{0}^{t}\frac{1}{2-3x}dx\;\;\;\ldots (2)\]
\[I_1=\lim_{t\rightarrow \frac{2}{3}^-}\left[\frac{-1}{3}\ln|2-3x|\right]_{0}^{t}\]
\[I_1=\lim_{t\rightarrow \frac{2}{3}^-}\left[\frac{-1}{3}\ln|2-3t|+\frac{1}{3}\ln 2\right]\]
$\;\;\;\;\;\;\;\;\;\;=$ does not exist
Since limit on R.H.S. of (2) does not exist so $I_1$ is divergent
From (1)
Consequently $I$ is divergent.
