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
