Proof that the improper integral $\frac{1}{(x^2-1)}$ from $0$ to $\infty$ is divergent.

Question

$$\int_0^\infty\frac{1}{x^2-1}$$

It was said that this integral is divergent. I have tried splitting the integral. $$\int_0^1\frac{1}{x^2-1}+\int_1^\infty\frac{1}{x^2-1}$$

Using this, I tried to prove $\int_0^1\frac{1}{x^2-1}$ is divergent by comparison test, but there are no functions that lie under $\frac{1}{x^2-1}$ that diverge (at least that I can think of).

I know the function diverges by Wolfram Alpha, but am struggling to prove it.

Answer 1

Notice that

$$\frac{1}{x^2 - 1} = \frac{1}{x + 1} \cdot \frac{1}{x - 1} \ge \frac 1 3 \frac{1}{x - 1}$$

for $1 \le x \le 2$. Now compute

$$\int_t^2 \frac{dx}{x - 1}.$$