Substitute #1+e^x=u^2#. Then #e^x dx = 2u\ du#. Thus the integral becomes
#int\ (e^x+sqrt(1+e^x))/(2+e^x-e^(2x))dx = int\ (u^2-1+u)/(2+(u^2-1)-(u^2-1)^2) (2u\ du)/(u^2-1)#
#qquad = int\ (2u(u^2+u-1))/((u^2-1)(-u^4+3u^2))\ du#
#qquad = int\ (2(u^2+u-1))/(u(u^2-1)(3-u^2))\ du#
We will now evaluate this integral by the method of partial fractions. Consider :
#(u^2+u-1)/(u(u^2-1)(3-u^2))= A/u+B/(u-1)+C/(u+1)#
#qquad qquad qquad qquad qquad qquad qquad qquad +D/(sqrt3-u)+E/(sqrt3+u)#
Thus
#u^2+u-1 = A(u^2-1)(3-u^2)+Bu(u+1)(3-u^2)#
#qquadqquadqquadquad +Cu(u-1)(3-u^2)+Du(u^2-1)(sqrt3+u)#
#qquadqquadqquadquad +Eu(u^2-1)(sqrt3-u)#
- Substituting #u=0#, we get
#-3A=-1 implies color(red)(A=1/3)#
- Substituting #u=1#, we get
#\ 4B=1 implies color(red)(B=1/4)#
- Substituting #u=-1#, we get
#4C=-1 implies color(red)(C=-1/4)#
- Substituting #u=sqrt3#, we get
#12D=2+sqrt3 implies color(red)(D=(2+sqrt3)/12)#
- Substituting #u=-sqrt3#, we get
#-12E=2-sqrt3 implies color(red)(E=-(2-sqrt3)/12)#
Thus
#(u^2+u-1)/(u(u^2-1)(3-u^2))= 1/(3u)+ 1/(4(u-1))-1/(4(u+1))#
#qquad qquad qquad qquad qquad qquad qquad qquad +(2+sqrt3)/(12(sqrt3-u))-(2-sqrt3)/(12(sqrt3+u))#
Hence our integral becomes
# int\ (2(u^2+u-1))/(u(u^2-1)(3-u^2))\ du = int[2/(3u)+ 1/(2(u-1))#
#qquad qquad -1/(2(u+1))+(2+sqrt3)/(6(sqrt3-u))-(2-sqrt3)/(6(sqrt3+u))]du#
#qquad = 2/3 log |u|+1/2 log|u-1|-1/2log|u+1|#
#qquadqquad -(2+sqrt3)/6 log|sqrt3-u|-(2-sqrt3)/6 log|sqrt3+u|+C#
#qquad = 2/3 log |u|+1/2 log|(u-1)/(u+1)|#
#qquadqquad -1/3log|3-u^2 |+sqrt3/6|(sqrt3+u)/(sqrt3-u)|#
Substituting #u# back in terms of #x#, the required integral becomes
#int\ (e^x+sqrt(1+e^x))/(2+e^x-e^(2x))dx#
#qquad =1/3 log(1+e^x)+1/2 log|(sqrt(1+e^x)-1)/(sqrt(1+e^x)+1)|#
#qquadqquad -1/3log|2-e^x|+sqrt3/6|(sqrt3+sqrt(1+e^x))/(sqrt3-sqrt(1+e^x))|+C#
