Given dy/dx=square root(-4x^2+8x+12), what is the function y?

The derivative of y is given as dy/dx =
sqrt(-4x^2+8x+12)

y' = sqrt [
-4x^2+8x+12]

=> 2*sqrt [-x^2 + 2x +
3]

=> 2*sqrt [ -(x^2 - 2x + 1) +
4]

=> 2*sqrt [ -(x - 1)^2 +
4]

y = Int[ y'  dx]

=>
2* Int [ sqrt [ 4 - (x - 1)^2] dx]

Let x - 1 = 2*sin
u

dx = 2*cos u du

2* Int [
sqrt [ 4 - (x - 1)^2] dx]

=>2*Int [ sqrt (4 - 4 (sin
u)^2) 2 cos u du]

=> 2*[Int[ 4*sqrt(1 - (sin u)^2)
cos u du]

=> 2*[Int[ 4*sqrt((cos u)^2) cos u
du]

=> 2*[Int[ 4*(cos u)^2
du]

=> 2*[Int[ 2*(1 + cos 2u)
du]

=> 2*[Int[ 2 + 2*cos 2u
du]

=> 2*2*u + 2*sin 2u +
C

=> 4u + 2*sin 2u +
C

=> 4*arc sin ((x - 1)/2) + 2*sin(2*arc sin ((x -
1)/2) + C

The function y = 4*arc sin ((x -
1)/2) + 2*sin(2*arc sin (x - 1)/2) + C