# Tricky integral

1. Jun 30, 2008

### fk378

1. The problem statement, all variables and given/known data
Integrate.
1/ [e^x (1-e^(-2x))^1/2]

2. Relevant equations
integration by parts

3. The attempt at a solution
First I split up the integral so it would be (1/e^-x)[(1/(1-e^-2x)^1/2]
Then I set u=e^-x, dv= (1-e^-2x)^-1/2
du= -e^-x
For my v I got 2(1-e^-2x)^1/2 but I don't know how to go about with the e^-2x, as in how to finish solving for the v-part.

I tried u-substitution and I get another term in the denominator. Then when I tried to differentiate the answer I originally got for v, I did not have anything in the denominator... can anyone help?

2. Jun 30, 2008

### rock.freak667

$$\int \frac{1}{e^x \sqrt{1-e^{-2x}}} dx$$

try $t^2=1-e^{-2x}$

3. Jun 30, 2008

### fk378

Is that valid? I just tried to solve it using your suggestion and with u=e^-x. With the u-subst I end up getting the arcsin integral, but I get something different with the t^2 substitution.

4. Jun 30, 2008

### rocomath

That's what I did too and verified it by taking it's derivative, but Mathematica gives a totally different solution.

I changed mine to:

$$\int\frac{e^{-x}}{\sqrt{1-(e^{-x})^2}}dx$$

Used:

$$u=e^{-x} \rightarrow \ln{u}=-x$$

5. Jun 30, 2008

### fk378

Oh interesting, why did you take the natural log? Where does that come into play?

6. Jun 30, 2008

### rocomath

I let:

$$u=e^{-x}$$ be my u-sub

$$-\frac{du}{u}=dx$$ be my du-sub