Answer
The given improper integral
$$
\int_{0}^{1} \frac{1}{2-3 x} d x
$$
is divergent.
Work Step by Step
$$
\int_{0}^{1} \frac{1}{2-3 x} d x
$$
Observe that the given integral is improper because $ f(x)= \frac{1}{2-3 x}$ has the vertical asymptote $x=\frac{2}{3} $. Since the infinite discontinuity occurs at the middle of the interval $ [0,1] $ at $x = \frac{2}{3} $, we must use part (c) of Definition 3 with $ c=\frac{2}{3} $:
$$
\int_{0}^{1} \frac{1}{2-3 x} d x =\int_{0}^{\frac{2}{3}} \frac{1}{2-3 x} d x + \int_{\frac{2}{3}}^{1} \frac{1}{2-3 x} d x
$$
where
$$
\begin{split}
\int_{0}^{2 / 3} \frac{1}{2-3 x} d x & =\lim _{t \rightarrow(2 / 3)-} \int_{0}^{t} \frac{1}{2-3 x} d x \\
& =\lim _{t \rightarrow(2 / 3)^{-}}\left[-\frac{1}{3} \ln |2-3 x|\right]_{0}^{t} \\
& =-\frac{1}{3} \lim _{t \rightarrow(2 / 3)-}[\ln |2-3 t|-\ln 2] \\
& =\infty
\end{split}
$$
so the integral
$$
\int_{0}^{2 / 3} \frac{1}{2-3 x} d x
$$
is divergent , so the given improper integral
$$
\int_{0}^{1} \frac{1}{2-3 x} d x
$$
is divergent.
