Integration Problem

[1Seraphim1]

with the integral of (4x^2)/sqrt(-4x^2 + 16x -9)

why does the denominator factor into sqrt(7 - (2x-4)^2) I don't remember why that is. I know how to complete the square,
but i dont remember the process for a problem with a negative leading coefficient. I've worked it out several different ways and I still haven't gotten it to factor that way. Can anyone help? Please?

[cberg22]

Generally when completing the square I factor out the coefficient of the x^2, but it hasn't been done in this case, so I didn't do it for my working out. I believe I haven't made a mistake but if I have, let me know:

-4x^2+16x-9
= -(4x^2-16x)-9
= (-(2x-4)^2+16)-9
= -(2x-4)^2+7
= 7-(2x-4)^2

[WallShadow]

I might be a tad bit late with this,

I don't know if this helps but you might be able to use one of the reverse trigonometric function rules:

derivative of arcsin u = u' / sqrt( 1 - u^2 )