How to prove the convergence of $\int_0^\infty \frac{\ln{x}}{1+x^2}dx$? How to prove the convergence of $\int_0^\infty \frac{\ln{x}}{1+x^2}dx$.Since it's unbounded on both sides, we need to prove the convergence of both $\int_0^1 \frac{\ln{x}}{1+x^2}dx$ and $\int_1^\infty \frac{\ln{x}}{1+x^2}dx$.
$\frac{\ln{x}}{1+x^2}\le \frac{x}{1+x^2}$ on $[1,\infty)$ and both of them are nonnegative, and $\int_1^\infty \frac{x}{1+x^2}dx=\frac12[\ln{|1+x^2|}]^\infty_1=\infty$. Apparently basic comparison does not work in this case. 
For $\int_0^1 \frac{\ln{x}}{1+x^2}dx$, the function$\frac{\ln{x}}{1+x^2}$ is not even nonnegative, we can't apply any comparision test. How are we supposed to know its convergence?

You say that the function is not even non-negative.  Have you studied its sign properly on $[0,1]$?  What is the matter about its sign?
The function is actually negative on $(0,1]$, so minus the function is positive, and now all comparison tests work just fine.
Since $1+x^2$ lies between $1$ and $2$, we have for $\epsilon\gt 0$:
$$0\leq \int_{\epsilon}^1 \frac{-\log x}{1+x^2} dx\leq \int_{\epsilon}^1 (-\log x) dx,$$ 
so the function $|f(x)|$ (where $f$ is your integrand) is integrable, and hence so is $f$.
