Separate variables:
#dy = (4x + (4x)/sqrt(16-x^2)) dx#
Integrate. On the right side, a #u# substitution can be helpful if the integral cannot be "eyeballed":
#int dy = int (4x + (4x)/sqrt(16-x^2)) dx#
#int dy = int 4x dx + int (4x)/sqrt(16-x^2) dx#
#int dy = int 4x dx + int (4x)(16 - x^2)^(-1/2) dx#
If we let #u = 16 - x^2#, then #du = -2x dx#, or conversely:
#2x dx = -du => 4x dx = -2 du#
Substituting and integrating:
#int dy = int 4x dx + int u^(-1/2) (-2du)#
#int dy = int 4x dx -2int u^(-1/2) du#
#y = 2x^2 - 4u^(1/2) + C#
Lastly, substitute back:
#y = 2x^2 - 4(16-x^2)^(1/2) + C = 2x^2 -4sqrt(16-x^2) + C#
