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