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

The integral of (x+2)/(sqrt(4x-4x^2)) is -1/2 sqrt(x(-x+1))+5/4 arcsin(2x-1)+C

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

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

- \frac{1}{2} x (- x + 1) + \frac{5}{4} arcsin (2 x - 1) + C

Solution steps

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

Expand

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

= \int \frac{x}{2 x - x ^{2}} + \frac{1}{x - x ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{2 x - x ^{2}} d x + \int \frac{1}{x - x ^{2}} d x

\int \frac{x}{2 x - x ^{2}} d x = \frac{1}{4} (- 2 x (- x + 1) + arcsin (2 x - 1))

\int \frac{1}{x - x ^{2}} d x = arcsin (2 x - 1)

= \frac{1}{4} (- 2 x (- x + 1) + arcsin (2 x - 1)) + arcsin (2 x - 1)

Simplify

\frac{1}{4} (- 2 x (- x + 1) + arcsin (2 x - 1)) + arcsin (2 x - 1) : - \frac{1}{2} x (- x + 1) + \frac{5}{4} arcsin (2 x - 1)

= - \frac{1}{2} x (- x + 1) + \frac{5}{4} arcsin (2 x - 1)

Add a constant to the solution

= - \frac{1}{2} x (- x + 1) + \frac{5}{4} arcsin (2 x - 1) + C

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What is the integral of (x+2)/(sqrt(4x-4x^2)) ?
The integral of (x+2)/(sqrt(4x-4x^2)) is -1/2 sqrt(x(-x+1))+5/4 arcsin(2x-1)+C