integrales de funciones trigonometricas hiperbolicas

[Wes Loewer]

If I were doing this integral by hand, I would do the following:
∫(1-tanh(x)^2) dx = ∫sech(x)^2 dx = tanh(x)+C

The Prime CAS gives the result as -2/((e^x)^2+1)+C

At first glance, these might look different, but since tanh(x) - 1 = -2/((e^x)^2+1), then the two expressions differ only by a constant. This means that the two expressions are both correct, they just have different integration constants.

If you look at other CAS's, you'll see different but equivalent results.
Maxima: 2/(e^(-2*x)+1)
Nspire: −2/(e^(2*x)+1)
WolframAlpha: tanh(x)

When my students say that they got a different answer than the textbook, I encourage them to graph both results. If they get the same graph but shifted up or down, then they can be reasonable certain that their answers are equivalent.