Question

求积分∫xe^x/(x+1)^2dx解题步骤

Answer 1

∫xe^x/(x+1)^2dx

=∫xe^xd[-1/(x+1)]

=xe^x·[-1/(x+1)] -∫-1/(x+1)d(xe^x)

=-xe^x/(x+1)+∫1/(x+1) ·(1+x)e^xdx

=-xe^x/(x+1)+e^x +C

=e^x/(x+1) +C

注：∫UdV+∫VdU=UV