how do u intergrate (x-2)^-3?

Answer 1

take t = x - 2

dx/dt = 1

int (x - 2)^-3dx = int( t^-3 )dx = int (t^-3) dx/dt dt = int(t^-3)dt =
= (t^-2)/-2 + c = -(t^-2)/2 + c = -(x -2)^-2/2 + c = 1/[2(x - 2)^2] + c

Answer 2

Notice that the derivative of (x - 2)^-2 would just be
-2(x - 2)^(-3). The derivative of (x-2) is just 1, so applying the chain rule doesn't change anything.

So to balance out that -2, the integral of (x - 2)^-3 should be
(-1/2)(x - 2)^(-2) + c. You can verify this by taking the derivative and getting back the original expression.