What is the integral of (2x-1)/(sqrt(4x-4x^2)) ?

The integral of (2x-1)/(sqrt(4x-4x^2)) is -sqrt(x(-x+1))+C

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integral of (2x-1)/(sqrt(4x-4x^2))

\int \frac{2 x - 1}{4 x - 4 x ^{2}} d x

- x (- x + 1) + C

Solution steps

\int \frac{2 x - 1}{4 x - 4 x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{4 u} d u

= \int - \frac{1}{4 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{4} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= - \frac{1}{4} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 4 x - 4 x^{2}

= - \frac{1}{4} \cdot 2 (4 x - 4 x^{2})^{\frac{1}{2}}

Simplify

- \frac{1}{4} \cdot 2 (4 x - 4 x^{2})^{\frac{1}{2}} : - x (- x + 1)

= - x (- x + 1)

Add a constant to the solution

= - x (- x + 1) + C

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What is the integral of (2x-1)/(sqrt(4x-4x^2)) ?
The integral of (2x-1)/(sqrt(4x-4x^2)) is -sqrt(x(-x+1))+C