Question

How do you integrate `int (e^x-e^-x)/(e^x+e^-x)dx`?

Answer 1

`log(e^x+e^(-x))+C`

Explanation:

Note that `d/(dx)(e^x+e^(-x))=e^x-e^(-x)` so

`(e^x-e^(-x))/(e^x+e^(-x))= d/(dx)log(e^x+e^(-x))` and finally

`int (e^x-e^-x)/(e^x+e^-x)dx = int d/(dx)log(e^x+e^(-x))dx=log(e^x+e^(-x))+C`