Hi

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.

Thanks