If f(x/(x+2)) = 2x+1, (x ∈ R, x ≠ -2), then ∫f(x)dx is equal to: (where C is a constant of integration)

Let x/(x+2) = y ⇒ x = yx + 2y ⇒ x(1-y) = 2y+4 ⇒ x = (2y+4)/(1-y)
This gives us f(y) = 2((2y+4)/(1-y)) + 1
So, we have f(x) = 2((2x+4)/(1-x)) + 1 = (3x+9)/(1-x) = -(3x+9)/(x-1) = -3(x+3)/(x-1) = -3(x-1+4)/(x-1) = -3(1 + 4/(x-1))
∫f(x)dx = ∫(-3 - 12/(x-1))dx = -3x -12loge|x-1| + c
Thus ∫f(x)dx = -3x - 12loge|1-x| + c
So, the correct answer is option B.