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Math Help - Indefinite integral error

  1. #1
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    Indefinite integral error

    So I did this indifinite integral one way and got the wrong answer. Then I did it another way (right side), and got the correct answer. Still, I can't figure out why the first approach didn't work. Would appreciate any explanation!


    Indefinite integral error-img201.jpg
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  2. #2
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    Re: Indefinite integral error

    Both methods are correct. Your mistake is in the third line, you just made a mistake in your integration.

    2\int \frac{-e^{-x}}{e^{-x}+1}dx=2ln(e^{x}+1)-2x + C

    With this correction, your other 2x vanishes and you are left with a similar expression.
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Indefinite integral error

    If you write:
    2x+2ln(1+e^(-x))+C
    as
    2(x+ln(1+e^(-x))+C
    =2(ln(e^x)+ln(1+e^(-x)))+C
    =2(ln((e^x).(1+e^(-x)))+C
    =2 ln(e^x+1)+C
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  4. #4
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    Re: Indefinite integral error

    \frac{2}{e^{-x} + 1} = \frac{2e^x}{1+e^x}

    That help?
    Last edited by RogueDemon; July 14th 2011 at 02:03 PM.
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: Indefinite integral error

    @ Gralla55:
    There's nothing wrong with your solutions.
    Also it's not necessary to use 'absolute value signs' because e^(-x)+1>0 for each x.

    @DivisionByZero:
    That's not a mistake because:
    2ln(e^x+1)-2x=2ln(e^x+1)-2ln(e^x)=2ln((e^x+1)/(e^x))=2ln(e^(-x)+1)
    So just the same as Gralla55 said.
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