It's not that obvious at first. But its easy to understand.

$\displaystyle \int\frac{x}{\sqrt{4-x^2}}\,dx$

Let $\displaystyle u=4-x^2\implies\,du=-2x\,dx$

Note that in the integral, we have $\displaystyle \int\frac{{\color{red}x}}{\sqrt{4-x^2}}\color{red}\,dx$, not $\displaystyle -2x\,dx$. To make the $\displaystyle -2x\,dx$ appear, multiply the integral by $\displaystyle \frac{-2}{-2}$ (equivalent to 1)

Thus, we get the following:

$\displaystyle \frac{-2}{-2}\int\frac{x}{\sqrt{4-x^2}}\,dx\implies {\color{red}\frac{-1}{2}}\int\frac{-2x}{\sqrt{4-x^2}}\,dx$.

Now make the substitutions:

$\displaystyle \frac{-1}{2}\int u^{-\frac{1}{2}}\,du$.

Hope this makes sense!