First we substitute:
#u=e^(2x)+1;e^(2x)=u-1#
#(du)/(dx)=2e^(2x);dx=(du)/(2e^(2x))#
#intsqrt(u)/(2e^(2x))du=intsqrt(u)/(2(u-1))du=1/2intsqrt(u)/(u-1)du#
Perform a second substitution:
#v^2=u;v=sqrt(u)#
#2v(dv)/(du)=1;du=2vdv#
#1/2intv/(v^2-1)2vdv=intv^2/(v^2-1)dv=int1+1/(v^2-1)dv#
Split using partial fractions:
#1/((v+1)(v-1))=A/(v+1)+B/(v-1)#
#1=A(v-1)+B(v+1)#
#v=1#:
#1=2B#, #B=1/2#
#v=-1#:
#1=-2A#, #A=-1/2#
Now we have:
#-1/(2(v+1))+1/(2(v-1))#
#int1+1/((v+1)(v-1))dv=int1-1/(2(v+1))+1/(2(v-1))dv=1/2[-ln(abs(v+1))+ln(abs(v-1))]+v+C#
Substituting back in #v=sqrt(u)#:
#1/2[-ln(abs(sqrt(u)+1))+ln(abs(sqrt(u)-1))]+sqrt(u)+C#
Substituting back in #u=1+e^(2x)#
#1/2[-ln(abs(sqrt(1+e^(2x))+1))+ln(abs(sqrt(1+e^(2x))-1))]+sqrt(1+e^(2x))+C#
