Prove that integral converges How can I prove that this integral converges?
$$\int\limits_1^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}\ dx$$
I tried to show that function $0 < \frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ is decreasing and is continious for $x$ in $(1,\infty)$. Furthermore the above integral converges when series $\sum_{1}^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ converges.
Are there better methods?

Near $x=1$, since $\ln x=\ln(1+(x-1))\sim x-1$, and $\sqrt{x^2-1}\sim\sqrt{x-1}$, you can compare your integral with $\frac{1}{\sqrt{x-1}}$.
Near $x=\infty$ (let me express in this way), since $\ln x \ll x^\alpha$ for every $\alpha>0$, and $(x-1)\sqrt{x^2-1}\sim x^{3/2}$, you can compare your integral with $\frac{1}{x^{3/2-\alpha}}$ for any $\alpha>0$ such that $\frac{3}{2}-\alpha>1$ (take $\alpha=1/4$ for instance).
