A nice substitution

$I=\displaystyle \int_0^\infty{\frac{\ln(x)}{x^2+1}dx}$

$=\displaystyle\int_\infty^0 \frac{\ln(\frac1u)}{\frac1{u^2}+1}(-\frac1{u^2})du$ (Put $x=\frac1u$ )

$=\displaystyle\int_0^\infty \frac{-\ln(u)}{1+u^2}du=-I=0$

$\begin{aligned} I & = \int_0^\infty \frac {\ln x}{x^2+1} dx & \small \color{#3D99F6} \text{Using }\int_0^\infty f(x) \ dx = \int_0^\infty \frac {f\left(\frac 1x\right)}{x^2}\ dx \\ & = \frac 12 \int_0^\infty \left(\frac {\ln x}{x^2+1} + \frac {\ln \frac 1x}{x^2\left(\frac 1{x^2}+1\right)} \right) dx \\ & = \frac 12 \int_0^\infty \left(\frac {\ln x}{x^2+1} - \frac {\ln x}{1+x^2} \right) dx \\ & = \boxed 0 \end{aligned}$

Let $x = \frac{1}{t}$ and thus $dx = -\frac{dt}{t^2}$ .

$\begin{aligned} I=\int_0^\infty{\frac{\ln(x)}{x^2+1}dx} &= \int_\infty^0{\frac{\ln\left(\frac{1}{t}\right)}{\left(\frac{1}{t}\right)^2+1} \left(-\frac{dt}{t^2}\right)}\\ &= -\int_\infty^0{\frac{\ln1 - \ln t}{\left(\frac{1}{t}\right)^2\cdot t^2+1\cdot t^2} dt} \\ &= \int_0^\infty{\frac{- \ln t}{1+t^2} dt} \\ &= \int_0^\infty{\frac{- \ln x}{1+x^2} dx} = -I \end{aligned}$

Because $I=-I$ , we have $\boxed{I=0}$ .