Math Help - starting point for this integral?

1. 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.

2. 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 ...