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