e2x+1e2x−1 can be simplified by dividing both numerator and denominator by ex:
e2x+1e2x−1=exe2x+ex1exe2x−ex1=ex+e−xex−e−x
Now the numerator is the derivative of the denominator.
Let t=ex+e−x
dxdt=ex−e−x
dt=(ex−e−x)dx
The integral becomes:
∫tdt
=log∣t∣+C
Substituting back t=ex+e−x:
=log∣ex+e−x∣+C