Help evaluating convergence of an integral with parameter $a$

Question

So I have this improper integral:

$$\int_a^{\infty}\frac{1}{x(x^2-1)^{6a}}dx$$

So I need to evaluate(find intervals for) convergence regarding parameter $a$.

So i know the following convergence rule:

$$\int_{{\,a}}^{{\,\infty }}{{\frac{1}{{{x^p}}}\,dx}}$$

For $a>0$ it is convergent for: $p>1$ and divergent for $p\leq1$.

So with this i'd br looking only at $a>0$ in my problem? And since i can evaluate:

$$\int_a^{\infty}\frac{1}{x(x^2-1)^{6a}}dx \leq \int_a^{\infty}\frac{1}{x^{13a}}dx$$ Should i just look at $13a > 1$ for convergence? Or am i doing something wrong.

Any help regarding my approach would be helpful.