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    Could someone please give me a hint on how to do this integral?

    2/(e^x+1) dx u=e^x+1

    I get to integrating 2/(u^2-u) but stuck from here
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    (Original post by sdgvrgverg)
    Could someone please give me a hint on how to do this integral?

    2/(e^x+1) dx u=e^x+1

    I get to integrating 2/(u^2-u) but stuck from here
    Partial fractions!
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    I don't do partial fractions till C4, this is a C3 question (AQA)
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    well, you get

    du = e^x dx

    which is du = (u -1) dx.

    or -(u-1) du = dx.

    _Kar.

    Hint:
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    The answer I (possibly incorrectly) reached was Ln (u) - 2u + C. That might help if you're looking for the method.
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    Yes that is incorrect, thanks for trying
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    (Original post by sdgvrgverg)
    I don't do partial fractions till C4, this is a C3 question (AQA)
    If you're going to use that substitution straight off, you're going to end up with partial fractions.

    The only other thing I can suggest is to multiply top and bottom by e^(-x), and then use a similar substitution, which will get you round the partial fractions issue.
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    I'll try that, thanks.
    Strange though, it's in the C3 section of the book and says to use the substitution u=e^x+1
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    (Original post by sdgvrgverg)
    I'll try that, thanks.
    Strange though, it's in the C3 section of the book and says to use the substitution u=e^x+1
    It's possible I'm missing something, and also possible there's an error in the book (won't be the first time, in either case).
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    I agree with the latter!
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    about the error in the book, I doubt you're missing something!
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    We've I = \int\frac{2}{e^x+1}\;{dx}. Letting u = e^x+1 gives \int\frac{2}{u(u-1)}\;{du} . You don't have to know the method of partial fractions to observe that 2 = 2u-2(u-1).
    Thus I= \int\frac{2u-2(u-1)}{u(u-1)}\;{du} = 2\int\frac{1}{u-1}\;{du}-2\int\frac{1}{u}\;{du} = 2\ln\left(u-1\right)-2\ln{u}+k = 2\ln{e^x}-2\ln\left(e^x+1\right)+k = 2x-\ln\left(e^x+1\right)+k.
 
 
 
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