Answer
Diverges
Work Step by Step
Since
$\int_{0}^{2} \dfrac{d x}{1-x}=\int_{0}^{1} \dfrac{d x}{1-x}+\int_{1}^{2} \dfrac{d x}{1-x} $
Now, $ \int_{0}^{1} \dfrac{d x}{1-x}=\lim\limits_{b \to 1^{-}}\int_{0}^{b} \dfrac{d x}{1-x}\\
=\lim\limits_{b \to 1^{-}}[-\ln (1-x)]_{0}^{b}\\
=\lim\limits_{b \to 1^{-}} [-\ln (1-b)-0] \\=\infty$
Thus, the integral diverges.