What is the integral of (2-2x)/(sqrt(10x-5x^2)) ?

The integral of (2-2x)/(sqrt(10x-5x^2)) is 2/(sqrt(5))sqrt(-x^2+2x)+C

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integral of (2-2x)/(sqrt(10x-5x^2))

\int \frac{2 - 2 x}{10 x - 5 x ^{2}} d x

\frac{2}{5} - x^{2} + 2 x + C

Solution steps

\int \frac{2 - 2 x}{10 x - 5 x ^{2}} d x

\frac{2 - 2 x}{10 x - 5 x ^{2}} = \frac{2}{5} \frac{1 - x}{2 x - x ^{2}}

= \int \frac{2}{5} \cdot \frac{1 - x}{2 x - x ^{2}} d x

Take the constant out:

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

= \frac{2}{5} \cdot \int \frac{1 - x}{2 x - x ^{2}} d x

Expand

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

= \frac{2}{5} \cdot \int \frac{1}{2 x - x ^{2}} - \frac{x}{2 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

= \frac{2}{5} (\int \frac{1}{2 x - x ^{2}} d x - \int \frac{x}{2 x - x ^{2}} d x)

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

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

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

Simplify

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

= \frac{2}{5} - x^{2} + 2 x

Add a constant to the solution

= \frac{2}{5} - x^{2} + 2 x + C

\int_{0}^{2} 2 π (x) (2 x - x^{2}) d x

\int \frac{1}{sin ^{2} ( 4 x )} d x

x \to + 4 lim (\frac{∣ x - 4 ∣}{4 - x})

n = 0 \sum \infty 4^{n} x^{n}

\frac{\partial}{\partial y} (4 x + 3 y)

What is the integral of (2-2x)/(sqrt(10x-5x^2)) ?
The integral of (2-2x)/(sqrt(10x-5x^2)) is 2/(sqrt(5))sqrt(-x^2+2x)+C