Evaluate the improper integral I=∫01ln(x)1−x2dxI = \int_0^1 \frac{\ln(x)}{\sqrt{1-x^2}} dxI=∫011−x2ln(x)dx.
−π2ln(2)-\frac{\pi}{2} \ln(2)−2πln(2)
−π4ln(2)-\frac{\pi}{4} \ln(2)−4πln(2)
−πln(2)-\pi \ln(2)−πln(2)
−π2-\frac{\pi}{2}−2π