How can I prove that $\int_1^2 \frac{\ln(x-1)}{\ln(x^2-1)}~\text{d}x$ diverges?

Question

I am looking at the following integral:

$$\int_1^2 \frac{\ln(x-1)}{\ln(x^2-1)}~\text{d}x$$

Which, according to WolframAlpha, diverges. The integrand has a vertical asymptote at $x=\sqrt2$ over which it changes sign but is not odd about $x=\sqrt2$. Here is a graph of the integrand:

I am not sure how to prove this integral diverges and am hoping the MSE integration wizards can help. I cannot evaluate it directly, it is not symmetric, and it is unclear what I could compare it to. So, how can I approach this problem?

Update 1: numerically, it seems $$\int_1^2 \frac{\ln(x-1)}{\ln(x^2-1)}~\text{d}x$$ diverges to $+\infty$.

Update 2: here is a screenshot of what I used in Mathematica to look at how the integral of this function behaves near $x= \sqrt2$:

Answer 1

Hint: Observe \begin{align} \int^{\sqrt{2}}_1 \frac{\log(x-1)}{\log(x^2-1)}\ dx > \int^{\sqrt{2}}_{\sqrt{2}-\varepsilon} \frac{\log(1+(x-2))}{\log(1+(x^2-2))}\ dx>\int^{\sqrt{2}}_{\sqrt{2}-\varepsilon} \frac{\frac{1}{2}(x-2)^2-(x-2)}{2-x^2}\ dx \end{align} for some sufficiently small $\varepsilon>0$. Choose $\varepsilon = .1$ should do the trick.

Answer 2

Indeed at $\sqrt{2}$, there is a problem.

By substituting $u = x-\sqrt{2}$

$\frac{\ln(u+\sqrt{2}-1)}{\ln(u^2+2\sqrt{2}u+1)} \sim \frac{\ln(\sqrt{2}-1)}{u^2+2\sqrt{2}u}$ ($u\rightarrow 0$)

$\sim \frac{\ln(\sqrt{2}-1)}{2\sqrt{2}u}$

It is of the form $\frac{1}{u}$ and $\int_0^1 \frac1u du$ is divergent.