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

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

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

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

arcsin (2 x - 1) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{2}{- 4 u ^{2} + 1} d u

= \int \frac{2}{- 4 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 \frac{1}{- 4 u ^{2} + 1} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= 2 \cdot \int \frac{1}{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 d v

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

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

= arcsin (2 x - 1)

Add a constant to the solution

= arcsin (2 x - 1) + C

9 - x^{2}, - 8 x

\int 5^{x} + 6 sinh (x) d x

\frac{d}{d x} (\frac{x}{x + 5})

\frac{( 1 - x ^{2} ) d y}{d x} = x e^{y}

(2 x - 9)^{4} (x^{2} + x + 1)^{5}

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