∫1+xx+e-xdx=
logx-e-x+C
logx+e-x+C
log1+xex+C
1+xe-x2+C
log1-xex+C
Explanation for Correct answer:
Finding the value for the given integral:
∫1+xx+e-xdx=∫1+xx+1exdx=∫1+xxex+1exdx=∫ex1+xxex+1dx
Solving this by using the substitution method
Let t=xex+1dt=xex+1.exdx∵ddxuv=u.dv+v.du
∫1+xx+e-xdx=∫dtt=logt+C=logxex+1+C
Hence, option (C) is the correct answer.