How I solve this integral?

Let

#I=int(e^x+sqrt(1+e^x))/(2+e^x-e^(2x))dx#

Simplify things with the substitution #e^x=u#:

#I=int(u+sqrt(1+u))/(u(2+u-u^2))du#

Integration is distributive:

#I=int1/(2+u-u^2)du+intsqrt(1+u)/(u(2+u-u^2))du#

For the first integral, factorize the denominator. For the second integral, apply the substitution #1+u=v^2#:

#I=int1/((1+u)(2-u))du+2int1/((v^2-1)(3-v^2))dv#

Apply partial fraction decomposition:

#I=1/3int(1/(1+u)+1/(2-u))du+int(1/(v^2-1)+1/(3-v^2))dv#

For the first integral, integrate term by term. For the second integral, apply the difference of squares:

#I=1/3(ln|1+u|-ln|2-u|)+int(1/((v-1)(v+1))+1/((sqrt3+v)(sqrt3-v)))dv#

Apply the distributive property of integrals and partial fraction decomposition:

#I=1/3ln|(1+u)/(2-u)|+1/2int(1/(v-1)-1/(v+1))dv+1/(2sqrt3)int(1/(sqrt3+v)+1/(sqrt3-v))dv#

Integrate term by term:

#I=1/3ln|(1+u)/(2-u)|+1/2(ln|v-1|-ln|v+1|)+1/(2sqrt3)(ln|sqrt3+v|-ln|sqrt3-v|)+C#

Simplify:

#I=1/3ln|(1+u)/(2-u)|+1/2ln|(v-1)/(v+1)|+1/(2sqrt3)ln|(sqrt3+v)/(sqrt3-v)|+C#

Reverse the substitutions:

#I=1/3ln|(1+e^x)/(2-e^x)|+1/2ln|(sqrt(1+e^x)-1)/(sqrt(1+e^x)+1)|+1/(2sqrt3)ln|(sqrt3+sqrt(1+e^x))/(sqrt3-sqrt(1+e^x))|+C#