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Thread: Indefinite Integral

  1. Jan 15th 2012, 11:01 AM #1
    jacks
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    Indefinite Integral

    \displaystyle \int \frac{x}{\sqrt{1-x^2.(x+\sqrt{1-x^2})^8}}dx
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  2. Jan 15th 2012, 11:05 AM #2
    Siron
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    Re: Indefinite Integral

    Take a look here:
    int x/sqrt(1-x^2(x+sqrt(1-x^2))^8) - Wolfram|Alpha
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  3. Jan 15th 2012, 08:05 PM #3
    Charney
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    Re: Indefinite Integral

    There seems to be a random dot ( . ) in your equation, indicating a potential typo. Is this supposed to be \int \frac{x}{\sqrt{(1-x^2)(x + \sqrt{1-x^2})^8}} dx or \int \frac{x}{\sqrt{1-x^2(x + \sqrt{1-x^2})^8}} dx?
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