Answer
Diverges
Work Step by Step
Since
$$\int_{0}^{2} \frac{d x}{1-x}=\int_{0}^{1} \frac{d x}{1-x}+\int_{1}^{2} \frac{d x}{1-x} $$
and
\begin{align*}
\int_{0}^{1} \frac{d x}{1-x}&=\lim _{b \rightarrow 1^{-}} \int_{0}^{b} \frac{d x}{1-x}\\
&=\lim _{b \rightarrow 1^{-}}[-\ln (1-x)]_{0}^{b}\\
&=\lim _{b \rightarrow 1^{-}}[-\ln (1-b)-0]=\infty
\end{align*}
then $$\int_{0}^{2} \frac{d x}{1-x}$$
also diverges