f = (asin(2x+1))^2 / (-x -x^2)^1/2
let u = asin(2x+1)
2x + 1 = sin(u)
2dx/du = cos(u) = sqrt(1 - sin^2(u)) = sqrt(1 - (2x+1)^2) = 2sqrt(-x - x^2)
dx = sqrt(-x - x^2) du
So, int(f) dx = int((asin(2x+1))^2 / (-x -x^2)^1/2 ) sqrt(-x - x^2) du
int(f) dx = int(u^2) du = (1/3)u^3 + c
Answer: (1/3)(asin(2x+1))^3 + const