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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- Sep 23rd 2011, 10:45 AMKumastarting 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. - Sep 23rd 2011, 10:58 AMSironRe: starting point for this integral?
Rewrite the denominator as:

$\displaystyle \sqrt{2x-x^2}=\sqrt{\frac{x^2-2x}{-1}}=\sqrt{\frac{(x-1)^2-1}{-1}}=\sqrt{1-(x-1)^2}$

Now, you should recognize an arcsin by using a handy substitution ...