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Math Help - starting point for this integral?

  1. #1
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    starting point for this integral?

    not really sure where to start on this. Basically trying to find the indefinite integral of

    3/(2x-x^2)^0.5

    can't really figure out what I should do here, I don't think I can substitute and ibp.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: starting point for this integral?

    Rewrite the denominator as:
    \sqrt{2x-x^2}=\sqrt{\frac{x^2-2x}{-1}}=\sqrt{\frac{(x-1)^2-1}{-1}}=\sqrt{1-(x-1)^2}
    Now, you should recognize an arcsin by using a handy substitution ...
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