\(\displaystyle \int_{0}^{\infty}\frac{ln(x)}{1+x^{2}}dx\)
That's because this particular one is not easily done by elementary means. This is a toughy to do using parts, substitution, etc.
If we look at the graph from 1 to infinity, it has the same area as from 0 to 1. 0 to 1 is below the x-axis and 1 to infinity is above the x-axis. They cancel one another out, leaving 0.
\(\displaystyle \int_{0}^{1}\frac{ln(x)}{x^{2}+1}dx\approx{-0.916}\)
\(\displaystyle \int_{1}^{\infty}\frac{ln(x)}{x^{2}+1}dx\approx{0.916}\)
