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

The integral of 1/(1-sqrt(x)) is -2(-1+sqrt(x)+ln|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/(1-sqrt(x))

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

- 2 (- 1 + x + ln 1 - x) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{- u + 1}{u} : - 1 + \frac{1}{u}

= - 2 \cdot \int - 1 + \frac{1}{u} 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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= - 2 (- u + ln ∣ u ∣)

Substitute back

u = 1 - x

= - 2 (- (1 - x) + ln 1 - x)

Simplify

= - 2 (- 1 + x + ln 1 - x)

Add a constant to the solution

= - 2 (- 1 + x + ln 1 - x) + C

3 x (y^{2} + 1) d x + y (x^{2} + 2) d y = 0

y^{^{'}} = \frac{x - 1}{y - 4}

\frac{\partial}{\partial y} (3 y^{2} + 3 x^{2} - 12 y)

y^{^{'}} = 3 e^{- 2 x}

f (x) = \frac{1}{x}, (1, 3)

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