What is the integral of 1/(sqrt(4-x)) ?

The integral of 1/(sqrt(4-x)) is -2sqrt(4-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(4-x))

\int \frac{1}{4 - x} d x

- 2 4 - x + C

Solution steps

\int \frac{1}{4 - x} d x

Apply u-substitution

: \int - \frac{1}{u} d u

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

Take the constant out:

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = 4 - x

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

Apply radical rule:

= - 2 4 - x

Add a constant to the solution

= - 2 4 - x + C

\frac{d}{d x} (x^{4} - 4 x^{3} + 6 x^{2} - 36 x + 27)

\frac{d}{d x} (\frac{1 + csc ( x )}{1 + sec ( x )})

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

\frac{d}{d x} (\frac{11 x}{2 + x ^{2}})

\int_{0}^{10} 49 - 9.8 x d x

What is the integral of 1/(sqrt(4-x)) ?
The integral of 1/(sqrt(4-x)) is -2sqrt(4-x)+C