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**

f(x) = sqrt(-4x^2+8x+12)

We'll factorize by -4 the expression under the square root:

-4x^2+8x+12 = -4(x^2 - 2x - 3)

We'll re-write x^2 - 2x - 3 = x^2 - 2x + 1 - 4 and we'll recognize the perfect square: (x^2 - 2x + 1)

x^2 - 2x + 1 - 4 = (x - 1)^2 - 4

Int f(x)dx = sqrt4* Int sqrt[4 - (x - 1)^2]dx

We'll replace x- 1 by t.

We'll differentiate both sides:

dx = dt

2*Int sqrt[4 - (t)^2]dt

We'll substitute t = 2 sin u

u = arcsin (t/2)

We'll differentiate both sides:

dt = 2cos udu

2* Int sqrt[4 - 4(sin u)^2]*2cos udu = 8*Int (cos u)^2du

8* Int (cos u)^2du = (8/2)* Int (1 + cos 2u)du

4*Int (1 + cos 2u)du = 4* [Int du + Int (cos 2u)du]

Int f(x)dx = 4u + 2*sin 2u + C

**Int f(x)dx = 4*arcsin [(x-1)/2] + 2*sin {2arcsin [(x-1)/2]} + C**