不知道你怎么化着化着就跑偏了。∫[xe^x/(x+1)²]dx=∫[(x+1)e^x-e^x]/(x+1)²dx=∫[(x+1)(e^x)'-(x+1)'e^x]/(x+1)²dx=∫[e^x/(x+1)]'dx=∫d[e^x/(x+1)]=e^x/(x+1) +C
