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Math Help - Integration involving e and derivative of ln

  1. #1
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    Integration involving e and derivative of ln

    Find the indefinite integral.

    I know you have to use U substitution U=1/x^5 but I get lost from there.


    Find the derivative of y with respect to x.

    I was told to use the product rule on this one but again I'm stuck.



    Use implicit differentiation to find dy / dx.

    Lost on how to find the derivative of ln(xy)

    thanks in advance for any help.
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  2. #2
    MHF Contributor chisigma's Avatar
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    The only you need in last question is to remember that \ln (xy) = \ln x + \ln y...

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    \chi \sigma
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  3. #3
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    for the first one u=1/x^5 then du=-5/x^6 dx so can you see it now?
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  4. #4
    MHF Contributor chisigma's Avatar
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    ... for the first one is...

    u= x^{-5} \rightarrow x=u^{-\frac{1}{5}} \rightarrow dx= -\frac{1}{5}\cdot u^{-\frac{6}{5}}\cdot du

    ... after that...

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    \chi \sigma
    Last edited by chisigma; May 6th 2009 at 11:00 PM. Reason: forgotten 'du'... sorry...
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  5. #5
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    I got du= -5/x^6dx. I assume you divide by -5 so you end up with -1/5du=1/x^6. so you end up with -1/5 e^u du. Dunno what to do from there.
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  6. #6
    MHF Contributor chisigma's Avatar
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    ... the substitution u=x^{-5} \rightarrow dx= -\frac{1}{5}\cdot u^{-\frac{6}{5}}\cdot du leads to the substitution...

    \int \frac{e^{x^{-5}}}{x^{6}}\cdot dx \rightarrow -\frac{1}{5}\cdot \int e^{u}\cdot du

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    \chi \sigma
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  7. #7
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    thanks for the help.



    just need help on that one.
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  8. #8
    MHF Contributor chisigma's Avatar
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    If f=u\cdot v then df= u\cdot dv + v\cdot du. So that...

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    \chi \sigma
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