How to evaluate if $\int_2^\infty {\frac{1}{\log(x)\cdot \sqrt{x^2+1}}}dx$ converges?

Question

I am asked to evaluate if $$\int_2^\infty {\frac{1}{\log(x)\cdot \sqrt{x^2+1}}}dx$$ converges.

How can that be done? Even Wolframalpha/Mathematica 8.0 does not return a value.

Can this be done with improper integral convergence tests I,II or III?

EDIT:

Improper integral convergence test I (I guess it is also called comparison test I, at least in my native language):

If there exists $g(x)$ such that $0\leq f(x)\leq g(x)\quad \forall x \in [a;=\infty)$

and $\int_a^\infty{g(x)dx}$ converges, then $\int_a^\infty{f(x)dx}$ also converges.

Improper integral convergence test II would be the ratio test.

Answer 1

Since $$\int_{2}^{M}\frac{dx}{x\log x}=\int_{\log 2}^{\log M}\frac{dt}{t}=\log\log M-\log\log 2$$ and for any $x\geq 2$: $$\frac{1}{\log(x)\cdot\sqrt{x^2+1}}\geq\frac{1}{x\log x\cdot\sqrt{\frac{5}{4}}}=\frac{2}{\sqrt{5}}\cdot\frac{1}{x\log x}$$ the integral is divergent.

Answer 2

Hint. First investigate $$\int_2^\infty \frac{dx}{x\log x}\ .$$