What is the integral of 1/((4-x)^2) ?

The integral of 1/((4-x)^2) is 1/(4-x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/((4-x)^2)

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

\frac{1}{4 - x} + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

= - (- \frac{1}{u})

Substitute back

u = 4 - x

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

Simplify

= \frac{1}{4 - x}

Add a constant to the solution

= \frac{1}{4 - x} + C

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

(ln (cos (x)))^{^{'}}

x \to 0 lim (\frac{4 tan ( x )}{x})

\frac{d}{d x} (\frac{1}{e ^{- x^{2}}})

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