One integral problem

**r2d2** · January 28th 2010, 03:37 PM

Hi,

I am stuck on one integral problem.

How should I find the integral of:

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

Thanks for your help.

**General** · January 28th 2010, 03:59 PM

Originally Posted by r2d2

Hi,

I am stuck on one integral problem.

How should I find the integral of:

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

Thanks for your help.

Hello

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

**r2d2** · January 28th 2010, 04:06 PM

Originally Posted by General

Hello

Start by multiplying the integrated function by $\frac{e^{-x}}{e^{-x}}$ to get:
$\int \frac{e^{-x}}{e^{-x}-1}dx$ .
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:
$ln$ abs(e^(-x)-1) + C

Is this correct?

**General** · January 28th 2010, 04:11 PM

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:
$ln$ abs(e^(-x)-1) + C

Is this correct?

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

**r2d2** · January 28th 2010, 04:18 PM

Originally Posted by General

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

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 $x-ln|e^{-x}-1| + C$ . Do you know where they could have gotten the x from? Or is that incorrect?

**General** · January 28th 2010, 04:31 PM

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 $x-ln|e^{-x}-1| + C$ . Do you know where they could have gotten the x from? Or is that incorrect?

No.
The answer in the calculator is $x-ln|e^{{\color{red}x}}-1|+C$
Its $e^x$ not $e^{-x}$ .
See this:
integrate [ e^[-x] ] / [ e^[-x] - 1 ] - Wolfram|Alpha
Check it please.
It is the same.
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 $\frac{e^{-x}}{e^{-x}}$ 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.

I know this maybe will confuse you.
Do the following steps:
1- Differentiate $x-ln|e^x-1|+C$ .
2- Make a common denominator to get a fraction with a minus sign in the numerator.
3- Multiply this fraction by $\frac{e^{-x}}{e^{-x}}$ .
4- take the minus sign as common factor from the denominator and cancel it with the another one.
5- GOT IT ?

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