intergral of √(1+e^2x)dx?

Answer 1

First make the substitution u=1+e^(2x) to get it into the form:

0.5Int ( √u / (u-1) ) du

Next, make the trig substitution u = sec^2 t

Answer 2

Alternatively, let u = √(1+e^2x) Then

u² = (1+e^2x)
u² - 1 = e^2x
2u du = 2 e^2x dx = 2(u² - 1) dx
[u/(u² - 1)] du = dx

and the integral becomes

∫ [ u²/(u² - 1) ] du = ∫ [ (u² - 1 + 1)/(u² - 1) ] du = ∫ [1 + 1/(u² - 1)] du

You'll need partial fractions on the last integral, but this is as painless as I can come up with.