Integration of $\frac{dx}{(6x-4x^2)^{1/2}}$ and completing square. Integrate $$\frac{dx}{\sqrt{6x-4x^2}}$$
While completing the square of $6x-4x^2$, I want to know where did $9/16$ come from in the following after taking $4$ out as the common factor $$-4\bigg(x^2 -\frac{3x}{2} +\frac{9}{16}\bigg) +\frac{9}{4}$$. I want to know the steps of completing the square.

$$-4x^2+6x = -4(x^2-\frac{3}{2}x)$$
$$ = -4\bigg(\big(x-\frac{3}{4}\big)^2-\frac{9}{16}\bigg) $$
$$ = -4\big(x-\frac{3}{4}\big)^2+\frac{9}{4}$$
Do you see it now?
