Evaluate the improper integral ∫01lnx1−x2dx\int_0^1 \frac{\ln x}{\sqrt{1-x^2}} dx∫011−x2lnxdx.
−π2ln2-\frac{\pi}{2} \ln 2−2πln2
π2ln2\frac{\pi}{2} \ln 22πln2
−πln2-\pi \ln 2−πln2
πln2\pi \ln 2πln2