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

The integral of 1/((x-2)(x+2)^2) is 1/16 ln|x-2|-1/16 ln|x+2|+1/(4(x+2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{4 ( x + 2 )} + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x - 2 ) ( x + 2 ) ^{2}} : \frac{1}{16 ( x - 2 )} - \frac{1}{16 ( x + 2 )} - \frac{1}{4 ( x + 2 ) ^{2}}

= \int \frac{1}{16 ( x - 2 )} - \frac{1}{16 ( x + 2 )} - \frac{1}{4 ( x + 2 ) ^{2}} d x

Apply the Sum Rule:

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

= \int \frac{1}{16 ( x - 2 )} d x - \int \frac{1}{16 ( x + 2 )} d x - \int \frac{1}{4 ( x + 2 ) ^{2}} d x

\int \frac{1}{16 ( x - 2 )} d x = \frac{1}{16} ln ∣ x - 2 ∣

\int \frac{1}{16 ( x + 2 )} d x = \frac{1}{16} ln ∣ x + 2 ∣

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

= \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{16} ln ∣ x + 2 ∣ - (- \frac{1}{4 ( x + 2 )})

Simplify

= \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{4 ( x + 2 )}

Add a constant to the solution

= \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{4 ( x + 2 )} + C

\int \frac{1}{x ^{2}} 1 - \frac{1}{x} d x

\frac{d y}{d x} = y cos (3 x + 2)

\frac{d y}{d t} + 7 y = e^{5 t}

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

x^{^{''}} + 3 x^{^{'}} + 2 x = 1

What is the integral of 1/((x-2)(x+2)^2) ?
The integral of 1/((x-2)(x+2)^2) is 1/16 ln|x-2|-1/16 ln|x+2|+1/(4(x+2))+C