How to prove this integral converge? $$\int_{1}^{\infty }\frac{\ln x}{1+x^2}\,{\rm d}x$$
So far i tried to use the comparison test with $\int_{1}^{\infty }\frac{\sqrt{x}}{1+x^2}$ but i noticed that it's not always true. any ideas?
 A: Note that 
$$\frac{\ln(x)}{1+x^2} < \frac{\ln(x)}{x^2} $$
and using integration by parts
$$\int_1^\infty\frac{\ln(x)}{x^2}dx = 1.$$
By comparison $\int_1^\infty\frac{\ln(x)}{1+x^2}dx$ converges.
A: As long as you can show that $\frac{\ln x}{\sqrt{x}}$ is bounded above on our interval, you can conclude convergence. And it is not hard to show that in fact $\lim_{x\to\infty}\frac{\ln x}{x^{1/2}}=0$.
Actually, in our interval, $\sqrt{x}$ is always $\gt \ln x$. Let $f(x)=x^{1/2}-\ln x$. We have $f(1)\gt 0$. By using the derivative, you can show that $f(x)$ reaches a minimum at $x=4$. Since $f(4)\gt 0$, it follows that $\sqrt{x}\gt \ln x$ for all $x$ in our interval. 
A: $\newcommand{\+}{^{\dagger}}
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$$
\int_{1}^{\infty}{\ln\pars{x} \over 1 + x^{2}}\,\dd x
=\overbrace{\bracks{{\int_{0}^{1}{\ln\pars{x} \over 1 + x^{2}}\,\dd x
+\int_{1}^{0}{\ln\pars{1/x} \over 1 + 1/x^{2}}\,\pars{-\,{\dd x \over x^{2}}}}}}
^{\ds{=\ 0}}\
+\
\bracks{\overbrace{-\int_{0}^{1}{\ln\pars{x} \over 1 + x^{2}}\,\dd x}
^{\ds{=\ G}}}
$$
$\ds{G \approx 0.9160}$ is the
Catalan Constant
and the remaining integral is a well known representation of $\ds{G}$ as reported in the above link.

$$
\color{#77f}{\large\int_{1}^{\infty}{\ln\pars{x} \over 1 + x^{2}}\,\dd x = G}
\approx 0.9160
$$

A: Make the change of variables $u=\ln(x)$
$$ \int_{1}^{\infty }\frac{\ln x}{1+x^2} =  \int_{0}^{\infty }\frac{ue^u}{1+e^{2u}}.$$
Now when $u$ close to zero the integrand behaves as $u/2$ which is integrable function. Try to finish the problem. 
Added: Here is another hint
$$ \frac{ue^u}{1+e^{2u}} \leq \frac{ue^u}{e^{2u}}= ue^{-u} $$
which is integrable function (you can use integration by parts) over $(0,\infty)$.
