What is the integral of 1/(sqrt(x)\sqrt{1-x)}dx ?

The integral of 1/(sqrt(x)\sqrt{1-x)}dx is 2arcsin(sqrt(x))+C

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integral of 1/(sqrt(x)\sqrt{1-x)}dx

integral

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

2 arcsin (x) + C

Solution steps

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

Apply u-substitution

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

Use the common integral:

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

= 2 arcsin (u)

Substitute back

u = x

= 2 arcsin (x)

Add a constant to the solution

= 2 arcsin (x) + C

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