proof that the series converges? I just need to make sure that I do it correctly 

Thanks in advance,
 A: Hint
$$
\frac{|\log x|}{1+x^2}\leq |\log x|
$$
A: You should be careful about how you simplify after substituting. $$\frac{e^u}{1+e^{2u}}=\frac{1}{e^{-u}+e^{u}}$$
Note also that $$\int_{-\infty}^0 ue^{-u}du$$ diverges. 
I can suggest another route: integrate by parts:
$$\int_0^1\frac{\log x}{1+x^2}dx=\left.\arctan x\log x\right|_0^1-\int_0^1\frac{\arctan x}{x}dx$$
Use that $$\frac{\arctan x}x\to 1$$
so that the integrand is bounded and continuous over $[0,1]$. You'll also have to show that $$\arctan x\log x\to 0 \text{ when } x\to 0$$
You get then that $$\int_0^1\frac{\log x}{1+x^2}dx=-\int_0^1\frac{\arctan x}{x}dx=-G$$
where $G$ is Catalan's constant.
A: You can use the fact $\ln(x) < x$ which makes it easier 
$$ \int_{0}^{1}\frac{\ln(x)}{1+x^2}dx <  \int_{0}^{1}\frac{x}{1+x^2}dx=\frac{1}{2}\ln(1+x^2)|_{x=0}^{1}=\dots. $$
Note: Just note that, $ \frac{\ln(x)}{1+x^2} $ is Riemann integrable (in the extended sense if you wish and $x\sim 0$ $\frac{\ln(x)}{1+x^2}\sim \ln(x)$ ) and by the property of the Riemann integral, for $f$ and $g$ Riemann integrable functions such that 
$$ f \leq g\quad  \forall x\in[a,b] \implies \int_{a}^{b} f(x)dx \leq \int_{a}^{b} g(x)dx, $$
then above answer is true 
Another approach: Note that the integrand has a singularity at $x=0$, so we have
$$ x\sim 0 \implies \frac{\ln(x)}{1+x^2} \sim \ln(x) $$
and using the fact 
$$ \int_{0}^{1}\ln(x)dx < \infty $$
the original integral converges.
