答：∫(1到e)lnxdx=(xlnx-x)|(1到e)=elne-e-ln1+1=1∫(1到e)(lnx)²dx=[x(lnx)²-2xlnx+2x]|(1到e)=e(lne)²-2elne+2e-(ln1)²+2ln1-2=e-2 ∫(1到e)(lnx)²dx