Examine $\int \limits_{0}^{\infty} \frac{\ln x}{1+x^{2}} d x$ for convergence by using the direct comparison test 
Examine the following improper integral for convergence and determine the value if necessary.
$$\int_{0}^{\infty} \frac{\ln x}{1+x^{2}} dx$$

My solution approach:
I am not sure how to use the comparison test properly. This is how I determined the value of the integral.
$$
\begin{split}
I &= \int_{0}^{+\infty} \frac{\ln x}{x^{2}+1} dx
     \quad \text{substitute } x=\frac{1}{t}, d x=\frac{-1}{t^{2}} d t\\
  &=\int_{+\infty}^{0} \frac{\ln \frac{1}{t}}{\frac{1}{t^{2}}+1}
                       \frac{-1}{t^{2}} dt \\
  &=\int_{+\infty}^{0} \frac{\ln t^{-1}}{\frac{1}{t^{2}}+1} \frac{-1}{t^{2}} dt\\
  &= \int_{+\infty}^{0} \frac{-\ln t}{t^{2}+1} \frac{-t^{2}}{t^{2}} d t\\
  &= \int_{+\infty}^{0} \frac{\ln t}{t^{2}+1} d t \\
  &= -\int_{0}^{+\infty} \frac{\ln t}{t^{2}+1} d t\\
  &= -\int_{0}^{+\infty} \frac{\ln x}{x^{2}+1} d x\\
  &\Longrightarrow I=-I \Longrightarrow I+I=0 \Longrightarrow 2 I=0 \Longrightarrow I=0
\end{split}
$$
I still need to show that it converges with the comparison test though. Thanks in advance!
 A: As a side-effect of your analysis, it's enough to check only one of $x\to0$ and $x\to\infty$ as they contribute equally.
For small $x$, then, what is an obviously good approximation for $\frac{1}{1+x^2}$ which happens to be an upper bound for it? 
A: For large $x$ is well known that $\ln x\leq\sqrt x$, then $\dfrac{\ln x}{1+x^2}\leq\dfrac{1}{x^{\frac{3}{2}}}$, so at infinity the integral converges.
For small $x$ we have $-\dfrac{\ln x}{1+x^2}\sim-\ln x$. Integral $-\displaystyle\int\limits_0^\delta\ln xdx$ can be calculated by integration by parts and it is finite.
A: We can use your idea:
$$\int\limits_0^{+\infty}\frac{\ln{x}}{1+x^2}dx=\int\limits_0^{1}\frac{\ln{x}}{1+x^2}dx+\int\limits_1^{+\infty}\frac{\ln{x}}{1+x^2}dx=$$
$$=\int\limits_0^{1}\frac{-\ln{\frac{1}{x}}}{1+\frac{1}{x^2}}\left(-d\frac{1}{x}\right)+\int\limits_1^{+\infty}\frac{\ln{x}}{1+x^2}dx=$$
$$=\int\limits_{+\infty}^1\frac{\ln{x}}{1+x^2}dx+\int\limits_1^{+\infty}\frac{\ln{x}}{1+x^2}dx=0.$$
$\int\limits_1^{+\infty}\frac{\ln{x}}{1+x^2}dx$ converges because there is $a>1$ for which:
$$\int\limits_a^{+\infty}\frac{\ln{x}}{1+x^2}dx<\int\limits_a^{+\infty}\frac{1}{x^{1.5}}dx.$$
