Fast calculation for $\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=0$ I want to show that $\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=0$, but is there a faster method than finding the contour and doing all computations?
Otherwise my idea is to do the substitution $x=e^t$, integral than changes to $\int _{-\infty}^{\infty}\frac{t e^t}{1+e^{2t}}dt$. Next step is to take the contour $-r,r,r+i\pi,-r+i\pi$ and integrate over it...

Use the substitution $x=\tan\theta$.
$$I=\int_0^{\pi/2} \ln(\tan\theta)\,d\theta=-\int_0^{\pi/2}\ln(\tan\theta)\,d\theta=0$$
