That's Plane Easy!

Just observe $I=\int_{0}^\infty \frac{\ln(x)}{1+x^2} dx = \int_{0}^{1} \frac{\ln(x)}{1+x^2} dx + \int_{1}^\infty \frac{\ln(x)}{1+x^2} dx= I_1+I_2$ Now making the substitution $y=\frac{1}{x}$ in $I_2$ , we have $I_2=-\int_{1}^{0} \frac{\ln(1/y)}{1+y^2} dy=-\int_{0}^{1} \frac{\ln(y)}{1+y^2} dy=-I_1$ Hence $I=I_1+I_2=0$

Choose a branch cut of natural log in the lower half of the complex plane. Take $\ln(z)=|z|$ for $z$ on the positive real ray.

Then

$\displaystyle\int_{-\infty}^{\infty}\frac{\ln(z)}{1+z^2}dz=\int_{-\infty}^{0}\frac{\ln(z)}{1+z^2}dz+\int_{0}^{\infty}\frac{\ln(z)}{1+z^2}dz$

$\displaystyle=\int_{-\infty}^{0}\frac{\ln(|z|)+\pi i}{1+z^2}dz +\int_{0}^{\infty}\frac{\ln(z)}{1+z^2}dz$

$\displaystyle=\int_{0}^{\infty}\frac{\pi i}{1+z^2}dz+2\int_{0}^{\infty}\frac{\ln(z)}{1+z^2}dz$ . $(\ast)$

Let $C$ be a large semicircle above the real axis. Since $\frac{\ln(z)}{1+z^2}|z|\to 0$ as $z$ becomes small, we may ignore the singularity at the origin by deforming $C$ among an arbitrarily small semicircle and apply Cauchy's integration formula to find find $\displaystyle\int_C\frac{\ln(z)}{1+z^2}dz=2\pi i \frac{\ln(i)}{2i}=\frac{\pi^2}{2}i$ .

Similarly, $\frac{\ln(z)}{1+z^2}|z|\to 0$ as $z$ becomes large, so as the radius of $C$ becomes large the arc does not contribute to the integral and we find

$\displaystyle\int_{-\infty}^{\infty}\frac{\ln(z)}{1+z^2}dz=\frac{\pi^2}{2}i$ . $(\ast \ast)$

Moreover,

$\displaystyle\int_{0}^{\infty}\frac{\pi i}{1+z^2}dz=\lim_{b\to\infty}\pi i(\tan^{-1}(b)-\tan^{-1}(0))=\frac{\pi^2}{2}i$ . $(\ast\ast\ast)$ .

Finally, combining $(\ast),(\ast\ast)$ and $(\ast\ast\ast)$ yields

$\displaystyle\frac{\pi^2}{2}i=\frac{\pi^2}{2}i+2\int_0^{\infty}\frac{\ln(z)}{1+z^2}dz$ ,

so

$\displaystyle\int_0^{\infty}\frac{\ln(z)}{1+z^2}dz=\boxed{0}$ .

$let\quad I=\int _{ 0 }^{ \infty } \frac { \ln { ( } x) }{ 1+x^{ 2 } } \, dx\\ put\quad u=\frac { 1 }{ x } \\ \Rightarrow I=\int _{ \infty }^{ 0 } \frac { \ln { ( } \frac { 1 }{ x } ) }{ 1+\left( \frac { 1 }{ x } \right) ^{ 2 } } \, \frac { -dx }{ x^{ 2 } } =-\int _{ 0 }^{ \infty } \frac { \ln { ( } x) }{ 1+x^{ 2 } } \, dx\\ I=-I\Rightarrow I=0$

Put $x= \tan \theta \implies {d}x= \sec^2 \theta {d}\theta$

$I= \int_0^{\infty} \frac{\ln(x)}{1+x^{2}} \, dx = \int_0^{\frac{\pi}{2}} \ln(\tan \theta) \, d\theta = \int_0^{\frac{\pi}{2}} \ln(\tan(\frac{\pi}{2} -\theta)) \, d\theta = \int_0^{\frac{\pi}{2}} \ln(\cot \theta) \, d\theta$ )

Adding yields $2I= \int_0^{\frac{\pi}{2}} \ln(\tan \theta)+ \ln (\cot\theta) \, d\theta$

$2I= \int_0^{\frac{\pi}{2}} \ln(\tan \theta \cdot \cot\theta) \, d\theta = \int_0^{\frac{\pi}{2}} \ln(1) \, d\theta=0$