How can I study the convergence of the improper integral $\int_{2}^{+\infty} \frac{dx}{x\ln^2x}$? I've tried applying Comparison and Limit Criteria (reducing it to some form of the $p$-series) to no avail. 
Rather than finding out this particular solution, I'd like to know how to even approach these kinds of exercises.
 A: $u=ln(x)$, $xdu=dx$, $\int_{ln(2)}^{+\infty}{1\over {u^2e^u}}e^udu$ converges
A: Notice that every integral of the form
$$
\int_{2}^\infty \dfrac{dx}{x \ln^p(x)}
$$
can be explicitly computed via substitution $u = \ln(x)$. 
A: Note that for $x\ge2$, $\frac{1}{x\ln^2(x)}>0$. One way you could try it is using the fact that:
$$\int_2^\infty\frac{1}{x^a}dx=\frac{2^{1-a}}{a-1},\,\,\,a>1$$
so if you can show that:
$$x\ln^2(x)>x$$
for $x\ge2$ then you can use this integral as a bound

As others have mentioned, you can also show that:
$$\int_2^\infty\frac{dx}{x\ln^2(x)}dx=\int_{\ln(2)}^\infty\frac{1}{u^2}du$$
which clearly has a finite value and can be calculated using the formula I gave above
A: Note that,
if $p > 0$ then
$\left(\dfrac1{\log^p(x)}\right)'
 = -p \dfrac{dx}{x\log^{p + 1}(x)}
$
so that
$\int \dfrac{dx}{x\log^{p + 1}(x)}
=-\dfrac1{p\log^p(x)}
$.
Therefore
$\int_a^{\infty} \dfrac{dx}{x\log^{p + 1}(x)}
=-\dfrac1{p\log^p(x)}|_a^{\infty}
=\dfrac1{p\log^p(a)}
$.
Your case is $p=1$.
Also note that
$\left(\dfrac1{\log(\log(x))}\right)'
 = -\dfrac1{x \log(x) \log^2(\log(x))}
$
so
$\int \dfrac1{x \log(x) \log^2(\log(x))}
=-\dfrac1{\log(\log(x))}
$.
See if you can establish what
$\left(\dfrac1{\log(\log(\log ... \log(x)))}\right)'
$
is
where there are
$n$ nested $\log$s.
