How to prove that: $\int_{-1}^{1} \frac{1}{x^2-1}dx$ diverges?

Partial fractions for sure (and logs will emerge), and consider the two improper integrals $$ \int_{-1}^0 \frac{1}{x^2-1} \ dx \ \text{ with } \int_0^1 \frac{1}{x^2-1} \ dx. $$

You need BOTH to converge for your integral to converge. Carry on...

Note that for $x\in (-1,1)$, $$\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}\leq \frac{1}{(-2)(x+1)}.$$ Hence $$\int_{-1}^{1} \frac{1}{x^2-1}dx\leq -\frac{1}{2}\int_{-1}^{1} \frac{1}{x+1}dx=-\left[\frac{\ln(x+1)}{2}\right]_{-1}^1=-\infty.$$

Here's a less orthodox answer. Consider just the interval $[0, 1)$, and divide it into intervals $[0, 1/2), [1/2, 2/3), \ldots, [1 - \frac{1}{n}, 1 - \frac{1}{n+1}), \ldots$, each of width $\frac{1}{n(n+1)}$. The value of $\frac{1}{1-x^2}$ at $x = 1 - \frac{1}{n}$ is $\frac{n^2}{2n-1}$. Multiplying widths by function values gives a lower Riemann sum $$\int_0^1 \frac{dx}{1-x^2} > \sum_{n=1}^\infty \frac{n}{(n+1) (2n-1)}$$ which diverges as the harmonic series.