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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

= \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

\int sin (v) cos (v) d v

\int sin (x - 4) d x

\int sin^{2} (x) e^{- 2 x} d x

\int - 3 x^{2} cos (x) d x

\int \frac{3 + 3 x - 2 x ^{2}}{5 - 7 x + 3 x ^{2} - x ^{3}} d x

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