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Math Help - Harder Integration

  1. #1
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    Harder Integration



    Thanks,
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  2. #2
    MHF Contributor Calculus26's Avatar
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    a )

    Recall

    log4(x) = ln(x)/ln(4)




    b) Differentiate the Rhs

    Keys tanh(x) ' = sech^2(x)

    sinh(x)cosh(x) = 1/2sinh(2x)
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  3. #3
    MHF Contributor chisigma's Avatar
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    \int_{0}^{12} \frac{\log_{4} (4+x)}{4+x}\cdot dx = \frac{1}{\ln 4}\cdot \int_{0}^{12} \frac{\ln (4+x)}{4+x}\cdot dx =

    = \frac{1}{2\cdot \ln 4}\cdot |\ln^{2}(4+x)|_{0}^{12} = \frac{3}{2}\cdot \ln 4

    Kind regards

    \chi \sigma
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  4. #4
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    Thanks but still don't understand

    Hi, thanks for your support guys but i still dont understand how u integration from step 2 to step 3
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  5. #5
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    Make the substitution a = ln(4+x), thus da = dx / (4+x)
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