I =int e^x/(x(e+1))dx = 1/((e+1))int (de^x)/(x)I=∫exx(e+1)dx=1(e+1)∫dexx
Perform integration by part to get
Let u=e^x; v = 1/xu=ex;v=1x
int (de^x)/x = e^x/x + int(de^x)/x^2∫dexx=exx+∫dexx2
Repeat the process of integration by parts,
int (de^x)/x = e^x/x + e^x/x^2 +int 2/x^3 de^x ∫dexx=exx+exx2+∫2x3dex
int (de^x)/x = e^x/x + e^x/x^2 +2/x^3e^x + 6/x^4e^x+24/x^5e^x + 120/x^5e^x+...
int (de^x)/x = e^x(1/x + 1/x^2 +2/x^3 + 6/x^4+...+((n-1)!)/x^n + ... )=
int (de^x)/x = e^x/x(1+ 1/x +2/x^2 + 6/x^3+...+ (n!)/x^n + ... )=E_i(x)
Therefore,
I = e^x/((e+1))(1/x + 1/x^2 +2/x^3 + 6/x^4+...+((n-1)!)/x^n + ... )
I = e^x/((e+1)x)(1+ 1/x +2/x^2 + 6/x^3+...+ (n!)/x^n + ... )
I = ( E_i(x))/((e+1))
where Ei is a form of differential integral.
