Find the integral?

Question

Find the integral of (2x-1)/(sqrt(4x-4x^2)) dx. 

 

Answer: (-1/2)*sqrt(4x-4x^2)+C

 

 

Answer 1

I = ∫(2x-1)/(sqrt(4x-4x^2)) dx

u = 4x - 4x^2

du = 4dx - 8xdx = -4(2x-1)dx

I = -1/4 ∫ 1/√(u) du = -1/4 ∫ u^(-1/2) du

I = -1/4 u^(1/2)/(1/2) + C

I = -1/2 √(4x - 4x^2) + C

check:

dI/dx = (-1/2)(1/2)(1/√(4x-4x^2))(4-8x)

= (2x-1)(1/√(4x-4x^2)) √