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

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

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integral of sqrt(x/(1-x))

\int \frac{x}{1 - x} d x

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

Solution steps

\int \frac{x}{1 - x} d x

Simplify

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

= \int \frac{x}{1 - x} d x

Apply u-substitution

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

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

Take the constant out:

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

= - 2 \cdot \int - u^{2} + 1 d u

Apply Trigonometric Substitution

: \int cos^{2} (v) d v

= - 2 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= - 2 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= - 2 \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

Apply the Sum Rule:

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

= - 2 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= - 2 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} = 3 - y

\int \frac{7 x ^{3} - 3 x ^{2} + 21 x + 6}{x ^{3} + 3 x} d x

\frac{d y}{d x} + 3 x^{2} y = sin (x) e^{- x^{3}}

\frac{d}{d x} (cos (sin (tan (2 π x))))

cos (x y) = 1 + sin (y)

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