just let and then we have
Integral, I = -4 * dx/(3+2x-x^2)^0.5
There is an integral sign between 4 and dx.
I = -4 * dx/((4-(x-1)^2)^0.5 <--- completing the square
Let x-1 = 2sin(#), where # is theta
The result is 4arcsin((x-1)/2) + c, where c is an arbitrary constant
Please explain how this works.