I=int_(1/e)^(e) (Lnx)^2018/(x+1)*dx
After using x=1/y and dx=-dy/y^2 transform,
I=int_(e)^(1/e) (Ln(1/y))^2018/(1/y+1)*(-dy/y^2)
=int_(e)^(1/e) (-Lny)^2018/((y+1)/y)*(-dy/y^2)
=int_(e)^(1/e) (Lny)^2018/(y^2+y)*(-dy)
=int_(1/e)^(e) (Lny)^2018/(y^2+y)*dy
=int_(1/e)^(e) (Lnx)^2018/(x^2+x)*dx
After summing 2 integrals,
2I=int_(1/e)^(e) (Lnx)^2018/(x+1)*dx+int_(1/e)^(e) (Lnx)^2018/(x^2+x)*dx
=int_(1/e)^(e) (x*(Lnx)^2018)/(x^2+x)*dx+int_(1/e)^(e) (Lnx)^2018/(x^2+x)*dx
=int_(1/e)^(e) ((x+1)*(Lnx)^2018*dx)/(x*(x+1))
=int_(1/e)^(e) ((Lnx)^2018*dx)/x
=[(Lnx)^2019/2019]_(1/e)^(e)
=2/2019
Thus,
I=int_(1/e)^(e) (Lnx)^2018/(x+1)*dx=1/2019