Evaluate integral: ∫0∞ln⁡x1+x2dx

In general
${\displaystyle I\left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{\ln x}{x^2+{\alpha }^2}dx}$
can be evaluated by using substitution ${\displaystyle u=\frac{{\alpha }^2}{x}\Rightarrow x=\frac{{\alpha }^2}{u}\Rightarrow dx=-\frac{{\alpha }^2}{u^2}du}$, then
${\displaystyle I\left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{\ln \left(\frac{{\alpha }^2}{u}\right)}{{\left(\frac{{\alpha }^2}{u}\right)}^2+{\alpha }^2}\cdot \frac{{\alpha }^2}{u^2}du}$
${\displaystyle ={\int }_0^{\mathrm{\infty }}\frac{2\ln \alpha -\ln u}{{\alpha }^2+u^2}du}$
${\displaystyle =2\ln \alpha {\int }_0^{\mathrm{\infty }}\frac{1}{{\alpha }^2+u^2}du-{\int }_0^{\mathrm{\infty }}\frac{\ln u}{u^2+{\alpha }^2}du}$
$=2\ln \alpha {\int }_0^{\mathrm{\infty }}\frac{1}{{\alpha }^2+u^2}du-I(\alpha )$
${\displaystyle I\left(\alpha \right)=\ln \alpha {\int }_0^{\mathrm{\infty }}\frac{1}{{\alpha }^2+u^2}du}$
The last integral can easily be evaluated since it is a common integral. Using substitution ${\displaystyle u=\tan \theta }$, the integral turns out to be
${\displaystyle I\left(\alpha \right)=\frac{\ln \alpha }{\alpha }{\int }_0^{\frac{\pi }{2}}d\theta }$
${\displaystyle =\frac{\pi \ln \alpha }{2\alpha }}$
Thus
${\displaystyle I\left(1\right)={\int }_0^{\mathrm{\infty }}\frac{\ln x}{x^2+1}dx=0}$

Here's another general result, consider for ${\displaystyle \left|a\right|\le 1}$:
${\displaystyle I(a,b)={\int }_0^{\mathrm{\infty }}\frac{x^a}{b^2+x^2}dx}$
${\displaystyle =\frac{b^{a-1}}{2}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{a-1}{2}}}{1+t}dt}$ (1)
${\displaystyle =\frac{b^{a-1}}{2}B(\frac{1+a}{2},\frac{1-a}{2})}$ (2)
${\displaystyle =\frac{b^{a-1}}{2}\frac{\pi }{\cos \left(\frac{\pi }{2}a\right)}}$ (3)
(1) is by subst. ${\displaystyle x=b\sqrt{t}}$
(2) is by the definition of Beta function: ${\displaystyle B(a,b)={\int }_0^{\mathrm{\infty }}\frac{x^{a-1}}{{(1+x)}^{a+b}}dx}$
(3) is by using ${\displaystyle B(a,b)=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }(a+b)}}$ and Euler's reflection formula
${\displaystyle \mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }(1-a)=\frac{\pi }{\sin \left(\pi a\right)}}$
Hence,
${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{x^a{\left(\log x\right)}^n}{b^2+x^2}dx=\frac{d^n}{d{a}^n}\left(\frac{b^{a-1}}{2}\frac{\pi }{\cos \left(\frac{\pi }{2}a\right)}\right)}$
and when ${\displaystyle b=1,n=1,a=0}$, the result is 0

Another general approach :
Consider
${\int }_0^{\mathrm{\infty }}\frac{x^{b-1}}{a^2+x^2}dx$ (1)
Rewrite (1) as
${\int }_0^{\mathrm{\infty }}\frac{x^{b-1}}{a^2+x^2}dx=\frac{1}{a^2}{\int }_0^{\mathrm{\infty }}\frac{x^{b-1}}{1+(\frac{x}{a}{)}^2}dx$ (2)
Putting $x=ay\Rightarrow dx=ady$ yields
${\int }_0^{\mathrm{\infty }}\frac{x^{b-1}}{a^2+x^2}dx=a^{b-2}{\int }_0^{\mathrm{\infty }}\frac{y^{b-1}}{1+y^2}dy$
where
${\int }_0^{\mathrm{\infty }}\frac{y^{b-1}}{1+y^2}dy=\frac{\pi }{2\sin (\frac{b\pi }{2})}$
Hence
${\int }_0^{\mathrm{\infty }}\frac{x^{b-1}}{a^2+x^2}dx=\frac{\pi }{2}\cdot \frac{a^{b-2}}{\sin (\frac{b\pi }{2})}$ (3)
Differentiating (3) with respect to b and setting b=1 yields
${\int }_0^{\mathrm{\infty }}\frac{d}{db}[\frac{x^{b-1}}{a^2+x^2}{]}_{b=1}dx=\frac{\pi }{2}\frac{d}{db}[\frac{a^{b-2}}{\sin (\frac{b\pi }{2})}{]}_{b=1}$
${\int }_0^{\mathrm{\infty }}\frac{\ln x}{x^2+a^2}dx=\frac{\pi \ln a}{2\alpha }$
Thus
${\int }_0^{\mathrm{\infty }}\frac{\ln x}{x^2+1}dx=0$