求定积分∫上限e下限1/elnx的绝对值dx

Question

如题所述

Answer 1

解：∫<1/e,e>│lnx│dx=∫<1/e,1>(-lnx)dx+∫<1,e>lnxdx
=(-1/e+∫<1/e,1>dx)+(e-∫<1,e>dx) (应用分部积分法)
=(-1/e+1-1/e)+(e-e+1)
=2-2/e
=2(1-1/e)。