Q:

How to prove that the improper integral $\int^{\infty}_{1}\frac{\ln{x}}{1+x^2}dx$ is convergent?

How to prove that the improper integral $\int^{\infty}_{1}\frac{\ln{x}}{1+x^2}dx$ is convergent?
I tried both basic and limit comparison but I couldn't find a proper measure. i.e. $\frac{\ln{x}}{1+x^2}\le \frac{x}{1+x^2}$, but $\frac{x}{1+x^2}$ diverge.

A:

$\ln x < 2\sqrt x$ for all $x>0$, hence
$$0\le \int_1^\infty\frac{\ln x}{1+x^2}\,\mathrm d\mkern1mu x\le\int_1^\infty\frac{\sqrt x}{1+x^2}\,\mathrm d\mkern1mu x\le\int_1^\infty\frac{\sqrt x}{x^2}\,\mathrm d\mkern1mu x=\int_1^\infty\frac1{x^{3/2}}\,\mathrm d\mkern1mu x,$$
which converges.

