Evaluate the integral I=∫0∞ln(x)1+x2dxI = \int_0^{\infty} \frac{\ln(x)}{1+x^2} dxI=∫0∞1+x2ln(x)dx.
000
π\piπ
−π28-\frac{\pi^2}{8}−8π2
ln(2)\ln(2)ln(2)