Question

How do you find the antiderivative of `{(e^x)/ [(e^(2x)) - 1]}`?

Answer 1

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

Explanation:

`I=inte^x/(e^(2x)-1)dx`

Let `u=e^x` so that `du=e^xdx`:

`I=inte^x/((e^x)^2-1)dx=int1/(u^2-1)du`

Now, we can ley `u=sectheta`. This implies that `du=secthetatanthetad theta`.

Plugging these in:

`I=int(secthetatantheta)/(sec^2theta-1)d theta`

Note that `1+tan^2theta=sec^2theta`, so `sec^2theta-1=tan^2theta`:

`I=int(secthetatantheta)/tan^2thetad theta`

`I=intsectheta/tanthetad theta`

`I=int1/costheta(costheta/sintheta)d theta`

`I=intcsctheta`

This is a fairly common integral:

`I=-ln(abs(csctheta+cottheta))`

We need to rewrite this using `u=sectheta`. This means we have a right triangle where `u` is the hypotenuse, `1` is the side adjacent to `theta`, and `sqrt(u^2-1)` is the side opposite `theta`.

Thus `csctheta=u/sqrt(u^2-1)` and `cottheta=1/sqrt(u^2-1)`.

So:

`I=-ln(abs(u/sqrt(u^2-1)+1/sqrt(u^2-1)))+C`

`I=-ln(abs((u+1)/sqrt(u^2-1)))+C`

Bringing in the negative `1` as an exponent through the rule `Blog(A)=log(A^B)`:

`I=ln(abs(sqrt(u^2-1)/(u+1)))+C`

Through `u=e^x`:

`I=ln(abs(sqrt(e^(2x)-1)/(e^x+1)))+C`

Notice that `sqrt(e^(2x)-1)>0` and `e^x+1>0` for all values of `x`, so the absolute value bars can be removed:

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