Question

How do you integrate `int e^(2x)/(1+e^(2x))dx`?

Answer 1

The answer is `=1/2ln(1+e^(2x))+C`

Explanation:

Let's do it by substitution

Let `u=1+e^(2x)`

then, `du=2e^(2x)dx`

`int(e^(2x)dx)/(1+e^(2x))`

`=1/2int(du)/u=1/2lnabsu`

`=1/2ln(1+e^(2x))+C`