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

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

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integral of sqrt(x/(2-x))

\int \frac{x}{2 - x} d x

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

Solution steps

\int \frac{x}{2 - x} d x

Apply u-substitution

: \int - \frac{- u + 2}{u} d u

= \int - \frac{- u + 2}{u} d u

Take the constant out:

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

= - \int \frac{- u + 2}{u} d u

Apply Integration By Parts

: \frac{- u + 2}{u} u - \int - \frac{1}{u ^{\frac{1}{2}} - u + 2} d u

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

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

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

Substitute back

u = 2 - x

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

Simplify

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

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

Add a constant to the solution

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

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