Question

`int(2e^x - 2e^-x) / (e^x+e^-x)^2 dx =` ?

Answer 1

`-sechx+C`

Explanation:

`sinhx=(e^x-e^(-x))/2` and `coshx=(e^x+e^(-x))/2` then

`int(2e^x - 2e^-x) / (e^x+e^-x)^2 dx = int tanhx/coshx dx = -sechx+C`