**Problem:** Decide whether the following integral converges:

**Solution:**
There is just one problem, the infinity, and the given function is positive
on the integration interval, so we can apply comparison tests to the given
integral. A quick glance shows that evaluation is very difficult if not
impossible, so the test approach is a wise decision. We note that if *x*
grows to infinity, then some parts of the expression can be ignored (in both
polynomials, only the highest power survives):

Justification:

In fact, we even have an inequality there, for
*x* > 2

So with a bit of luck, the Comparison test would work, too. It remains to investigate the test integral. One possibility is a direct computation, first one needs to find an antiderivative using standard integration by parts:

If you don't like working with the logarithm, you can get rid of it by substitution, then apply integration by parts:

Thus

The test integral converges, hence by the Comparison test (or the Limit Comparison test), the given integral also converges.

**Note:** There is a way to avoid integrating the test function. We
observed that logarithm grows slower than any power. Since we have a square
in the denominator, we can afford to lose a little bit of it and still
obtain a convergent test integral. Precisely, using a limit argument one can
show that there is a certain *K* > 2*x* > *K**x* > *K*

We now obtained a different and simpler test function. Since we know that the
integral of *h* from *K* to infinity converges, by the Comparison
Test (and also noting that there are no problems between 2 and *K*) we
conclude that the given integral converges.