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...

If you use the substitution $x=1/t$, you can check directly that
$$
\int_0^1\frac{\log x}{1+x^2}\,dx=-\int_1^\infty\frac{\log t}{1+t^2}\,dt.
$$
