∫(1+x−x−1) ex+x−1dx =
(x+1)ex+x−1+c
(x−1)ex+x−1+c
12xex+x−1+c
x.ex+x-1+c
∫ex+x−1(1+x−x−1)dx
=∫[x[1−1x2]+1]ex+x−1dx since ∫efxf'xgx+g'xdx=efxgx
=xex+x−1+c