Determine the value of I=∫01ln(1+x)1+x2dxI = \int_{0}^{1} \frac{\ln(1+x)}{1+x^2} dxI=∫011+x2ln(1+x)dx.
π8ln2\frac{\pi}{8} \ln 28πln2
π4ln2\frac{\pi}{4} \ln 24πln2
ln2\ln 2ln2
π212\frac{\pi^2}{12}12π2