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Math Help - One integral problem

  1. #1
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    One integral problem

    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.
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  2. #2
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    Quote Originally Posted by r2d2 View Post
    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?
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  3. #3
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    Quote Originally Posted by General View Post
    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:
    lnabs(e^(-x)-1) + C

    Is this correct?
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  4. #4
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    Quote Originally Posted by r2d2 View Post
    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:
    lnabs(e^(-x)-1) + C

    Is this correct?
    there is a small mistake.
    It should be {\color{red} -}ln|e^{-x}-1| + C.
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  5. #5
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    Quote Originally Posted by General View Post
    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?
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  6. #6
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    Quote Originally Posted by r2d2 View Post
    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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