∫xex1+x2dx=
ex+c
−exx+12+c
exx+12+c
exx+1+c
I=∫xex1+x2dx
=∫exx+1−1(x+1)2dx∵∫exf(x)+f′(x)dx=exf(x)+cI=∫ex1x+1−1(x+1)2dx=ex1x+1+c=exx+1+c