Is this a simple negative exponent, power chain rule
Or an inverse sin integral?
And if it's an inverse sin integral, how do i get rid of the first x under the radical?
$\displaystyle \int \frac{dx}{\sqrt{3x-x^2}} $
it's a pain-in-the-### arcsin ...
$\displaystyle 3x - x^2 = $
$\displaystyle -\left(x^2 - 3x + \frac{9}{4} - \frac{9}{4}\right) =
$
$\displaystyle -\left[\left(x - \frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2\right] =$
$\displaystyle \left(\frac{3}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2$
now use the general form of the integral ...
$\displaystyle \int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin\left(\frac{u}{a}\right) + C$