Evaluate $\int_0^\infty\frac{\ln x}{1+x^2}dx$ Evaluate $$\int_0^\infty\frac{\ln x}{1+x^2}\ dx$$
I don't know where to start with this so either the full evaluation or any hints or pushes in the right direction would be appreciated.  Thanks.

Here's another general result, consider for $|a|\le 1$:
\begin{align*}
  I(a,b) &= \int_{0}^{\infty} \, \frac{x^a}{b^2+x^2}\, dx \\
  &= \frac{b^{a-1}}{2} \int_{0}^{\infty} \, \frac{t^{(a-1)/2}}{1+t}\, dt \tag{1}\\
  &= \frac{b^{a-1}}{2}\, \mathrm{B}\left(\frac{1+a}{2},\frac{1-a}{2}\right) \tag 2\\
  &= \frac{b^{a-1}}{2}\, \frac{\pi}{\displaystyle \cos{\left(\frac{\pi}{2}a\right)}} \tag 3
\end{align*}
$(1)$ is by subst. $\displaystyle x=b\sqrt{t}$
$(2)$ is by the definition of Beta function: $\displaystyle \mathrm{B}(a,b)=\int_{0}^{\infty} \, \frac{x^{a-1}}{(1+x)^{a+b}} \, dx$
$(3)$ is by using $\displaystyle \mathrm{B}(a,b)=\frac{\Gamma{(a)}\Gamma{(b)}}{\Gamma{(a+b)}}$ and Euler's reflection formula $\displaystyle \Gamma{(a)}\Gamma{(1-a)}=\frac{\pi}{\sin{\displaystyle \left({\pi}\, a\right)}}$
Hence, 
\begin{align*}
  \int_{0}^{\infty} \, \frac{x^a \left(\log{x}\right)^n}{b^2+x^2}\, dx &= \frac{\partial^{n} }{\partial a^n} \left(\frac{b^{a-1}}{2}\, \frac{\pi}{\cos{\displaystyle \left(\frac{\pi}{2}a\right)}}\right)
\end{align*}
and when $b=1, n=1, a= 0$, the result is $0$
Update: 
An even better result:
\begin{align*}
  \int_{0}^{\infty} \, \frac{x^a \left(\log{x}\right)^n}{b^c+x^c}\, dx &= \frac{\partial^{n} }{\partial a^n} \left(\frac{b^{a+1-c}}{c}\, \frac{\pi}{\sin{\displaystyle \left(\frac{1+a}{c}\pi\right)}}\right)
\end{align*}
where $\displaystyle 0<\frac{1+a}{c}<1$
