Q: $\displaystyle \displaystyle\int \frac{x}{\sqrt{4-x^2}} \, \mathrm{d}x$
Does that just equal $\displaystyle \sqrt{4-x^2}$? If so, why?
Let $\displaystyle u = 4 - x^2$
Then $\displaystyle \frac{du}{dx} = -2x$
$\displaystyle \int \frac{x}{\sqrt{4-x^2}} dx$
$\displaystyle = -\frac{1}{2} \int \frac{1}{\sqrt{u}} du$
$\displaystyle = -\frac{1}{2} \int (u)^{- \frac{1}{2}} du$
$\displaystyle = -\frac{1}{2} (2(u)^{\frac{1}{2}})$
$\displaystyle = -(\sqrt{4 - x^2})$