I want to find $p$ so that the integral converges.
$$\int_{0}^{1} \frac{1}{x^p(x^2-1)}dx$$
I know when $p=0$, the integral is divergent. I already tried for several values of $p$, like $p=1, p=2, p=-1$. As a result, the integral diverges.
In my opinion, there are no values of $p$ such that the integral converges. But how can I prove it? Any idea? Thanks in advance.
Multiple ways to prove this I think. One way, note that I flipped the sign $(x^2-1) \to (1-x^2)$ (if one converged, the other would as well):
$$ \int_0^1 \frac{dx}{x^p(1-x^2)} \ge \int_0^1 \frac{dx}{1-x^2} = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)\bigg |_{x=0}^{x=1} \to \infty. $$