Evaluate ∫01ln(x+1)1 dx=∫01ln(x+1) dx\int_0^1 \frac{\ln(x+1)}{1}\,dx = \int_0^1 \ln(x+1)\,dx∫011ln(x+1)dx=∫01ln(x+1)dx.
2ln2−12\ln 2 - 12ln2−1
ln2−1\ln 2 - 1ln2−1
ln2\ln 2ln2
1−ln21 - \ln 21−ln2