I was evaluating $\int_{-1}^{1}\frac{dx}{1-x^2}$ and came to the conclusion it does not converge. I understand it is because its primitive, $F(x)=\frac{1}{2}[\ln(1+x)-\ln(1-x)]$, does not converge at either endpoint. At the same time, $\int_{-1}^{1}\frac{dx}{\sqrt{1-x^2}}$ converges to $\pi$, which is verified through a simple variable change $x=\cos{u}\Rightarrow\int_{-1}^{1}\frac{dx}{\sqrt{1-x^2}}=\int_{0}^{\pi}\frac{\sin{u}}{|\sin{u}|}du$.
Is it possible to determine whether or not an improper integral between two assymptotes will converge or not, without having to find its primitive?