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

The integral of 1/(sqrt(1-\sqrt{x))} is -4(-1/3 (1-sqrt(x))^{3/2}+sqrt(1-\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(1-\sqrt{x))}

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

- 4 (- \frac{1}{3} (1 - x)^{\frac{3}{2}} + 1 - x) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Sum Rule:

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

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

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

= - 4 (- \frac{u ^{3}}{3} + u)

Substitute back

u = 1 - x

= - 4 - \frac{( 1 - x ) ^{3}}{3} + 1 - x

Simplify

- 4 - \frac{( 1 - x ) ^{3}}{3} + 1 - x : - 4 (- \frac{1}{3} (1 - x)^{\frac{3}{2}} + 1 - x)

= - 4 (- \frac{1}{3} (1 - x)^{\frac{3}{2}} + 1 - x)

Add a constant to the solution

= - 4 (- \frac{1}{3} (1 - x)^{\frac{3}{2}} + 1 - x) + C

\int \frac{x + 3}{( x + 1 ) ( x ^{2} + x + 3 )} d x

\int - \frac{1}{e ^{x} + 1} d x

\int \frac{6 x ^{2} + 1}{x ^{2} ( x - 1 ) ^{3}} d x

\int cos (1 - 3 x) d x

\int \frac{x ^{4} + x - 3}{x ^{3}} d x

What is the integral of 1/(sqrt(1-\sqrt{x))} ?
The integral of 1/(sqrt(1-\sqrt{x))} is -4(-1/3 (1-sqrt(x))^{3/2}+sqrt(1-\sqrt{x)})+C