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Math Help - Difficult Integral

  1. #1
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    Difficult Integral

    Evaluate this integral:

    \int \frac{6x+4}{\sqrt(24x - 9 - 4x^2)} dx


    Following an example in my texbook, I've split the top line:

    I need a substitution u=24x - 9 - 4x^2 where \frac{du}{dx} = 24 - 8x .
    Hence  6x+ 4 \equiv 14x - 20 + (24 - 8x).

    Also, completing the square  24x - 9 - 4x^2 = -4(x-3)^2 + 27

    So the integral can be written as:

     \int \frac{14x}{\sqrt(24x - 9 - 4x^2)} dx - 20\int \frac{dx}{\sqrt(24x - 9 - 4x^2)} + \int \frac{24-8x}{\sqrt(24x - 9 - 4x^2)} dx

    I know I'll end up with an inverse trig function somewhere but that's where I get stuck.
    Could someone please explain what to do. Thanks a lot.
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  2. #2
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    Try something like

    \frac{-6}{8}(-8x+24)+22=6x+4 and then you can break it in two integrals and you will find a arcsin function
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