1. ## One integral problem

Hi,

I am stuck on one integral problem.

How should I find the integral of:

$\displaystyle \int\frac{dx}{1-e^x}\,$

2. Originally Posted by r2d2
Hi,

I am stuck on one integral problem.

How should I find the integral of:

$\displaystyle \int\frac{dx}{1-e^x}\,$

Hello

Start by multiplying the integrated function by $\displaystyle \frac{e^{-x}}{e^{-x}}$ to get:
$\displaystyle \int \frac{e^{-x}}{e^{-x}-1}dx$.
It is easier now, right?

3. Originally Posted by General
Hello

Start by multiplying the integrated function by $\displaystyle \frac{e^{-x}}{e^{-x}}$ to get:
$\displaystyle \int \frac{e^{-x}}{e^{-x}-1}dx$.
It is easier now, right?
That does make it easier. Could you check me?

I let u = e^-x, and then solving for the integral through this "u" substitution, I came out with:
$\displaystyle ln$abs(e^(-x)-1) + C

Is this correct?

4. Originally Posted by r2d2
That does make it easier. Could you check me?

I let u = e^-x, and then solving for the integral through this "u" substitution, I came out with:
$\displaystyle ln$abs(e^(-x)-1) + C

Is this correct?
there is a small mistake.
It should be $\displaystyle {\color{red} -}ln|e^{-x}-1| + C$.

5. Originally Posted by General
there is a small mistake.
It should be $\displaystyle {\color{red} -}ln|e^{-x}-1| + C$.
Oh, I see. The negative comes from taking the derivative of e^-x. I was curious; as I was checking my answer using an online integral calculator, they had the answer as It should be $\displaystyle x-ln|e^{-x}-1| + C$. Do you know where they could have gotten the x from? Or is that incorrect?

6. Originally Posted by r2d2
Oh, I see. The negative comes from taking the derivative of e^-x. I was curious; as I was checking my answer using an online integral calculator, they had the answer as It should be $\displaystyle x-ln|e^{-x}-1| + C$. Do you know where they could have gotten the x from? Or is that incorrect?
No.
The answer in the calculator is $\displaystyle x-ln|e^{{\color{red}x}}-1|+C$
Its $\displaystyle e^x$ not $\displaystyle e^{-x}$.
See this:
integrate [ e^[-x] ] / [ e^[-x] - 1 ] - Wolfram|Alpha
If you differentiate this answer and make a common denominator (here you will get a minus sign in the numerator of a fraction) and multiply the resulting fraction by $\displaystyle \frac{e^{-x}}{e^{-x}}$ and take the minus sign as common factor from the denominator and cancel it with the another minus sign, you will get the same function which we integrated it.
1- Differentiate $\displaystyle x-ln|e^x-1|+C$.
3- Multiply this fraction by $\displaystyle \frac{e^{-x}}{e^{-x}}$.