Investigate the convergence of $\int_1^\infty \frac{\cos x \ln x}{x\sqrt{x^2-1}} \, dx$ 
Investigate the convergence of $$\int_1^\infty \frac{\cos x \ln x}{x\sqrt{x^2-1}} \, dx$$

so first of all let's split the integral to:
$$I_1 = \int_1^2 \frac{\cos x \ln x}{x\sqrt{x^2-1}} \, dx, \hspace{10mm} I_2 = \int_2^\infty \frac{\cos x \ln x}{x\sqrt{x^2-1}} \, dx$$
I already showed that the limit of the integrand is $0$ when $x\to 1^+$. Therefore, $I_1$ converges since it's integrand is a continuous function. 
Regarding to $I_2$. At first I tried to prove it converges absolutely (eliminating the $\cos x$) but it didn't workout. Anyway, I think it's diverges but I don't know how to demonstrate it.

For large enough $x$ you have $$\left| \frac{\cos x \ln x}{x^2} \right| \le \frac{\ln x}{2 x^2}.$$ You should have little difficulty showing that $$\int_1^\infty \frac{\ln x}{x^2} \, dx < \infty.$$
