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Math Help - Small indefinite integral mistake

  1. #1
    s3a
    s3a is offline
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    Small indefinite integral mistake

    Here is what wolfram alpha says:
    http://www.wolframalpha.com/input/?i...al+2%2F(xln(8x))

    and the attached pdf file is what I did.

    What I do not get is why you don't divide it by 8 like I multiplied by x for the du/dx algebraic manipulation.

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    \frac{\mathrm{d} \ln (8x)}{\mathrm{d} x}\neq \frac{8}{x}

    \frac{\mathrm{d} \ln (8x)}{\mathrm{d} x}=\frac{{u}'}{u}
    where u=8x
    Thus,

    \frac{\mathrm{d} \ln (8x)}{\mathrm{d} x}=\frac{8}{8x}=\frac{1}{x}

    You might want to try it again with that.


    The reason that \frac{\mathrm{d} \ln (8x)}{\mathrm{d} x} is the same as \frac{\mathrm{d} \ln (x)}{\mathrm{d} x}
    is because \ln(8x)=\ln(8)+\ln(x) due to the properties of logs. So ln(8x) is a constant added to ln(x)
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