What is the integral of 1/((16-x^2)^2) ?

The integral of 1/((16-x^2)^2) is 1/256 ln|x+4|-1/(64(x+4))-1/256 ln|x-4|-1/(64(x-4))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((16-x^2)^2)

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

\frac{1}{256} ln ∣ x + 4 ∣ - \frac{1}{64 ( x + 4 )} - \frac{1}{256} ln ∣ x - 4 ∣ - \frac{1}{64 ( x - 4 )} + C

Solution steps

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

Take the partial fraction of

\frac{1}{( 16 - x ^{2} ) ^{2}} : \frac{1}{256 ( x + 4 )} + \frac{1}{64 ( x + 4 ) ^{2}} - \frac{1}{256 ( x - 4 )} + \frac{1}{64 ( x - 4 ) ^{2}}

= \int \frac{1}{256 ( x + 4 )} + \frac{1}{64 ( x + 4 ) ^{2}} - \frac{1}{256 ( x - 4 )} + \frac{1}{64 ( x - 4 ) ^{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}{256 ( x + 4 )} d x + \int \frac{1}{64 ( x + 4 ) ^{2}} d x - \int \frac{1}{256 ( x - 4 )} d x + \int \frac{1}{64 ( x - 4 ) ^{2}} d x

\int \frac{1}{256 ( x + 4 )} d x = \frac{1}{256} ln ∣ x + 4 ∣

\int \frac{1}{64 ( x + 4 ) ^{2}} d x = - \frac{1}{64 ( x + 4 )}

\int \frac{1}{256 ( x - 4 )} d x = \frac{1}{256} ln ∣ x - 4 ∣

\int \frac{1}{64 ( x - 4 ) ^{2}} d x = - \frac{1}{64 ( x - 4 )}

= \frac{1}{256} ln ∣ x + 4 ∣ - \frac{1}{64 ( x + 4 )} - \frac{1}{256} ln ∣ x - 4 ∣ - \frac{1}{64 ( x - 4 )}

Add a constant to the solution

= \frac{1}{256} ln ∣ x + 4 ∣ - \frac{1}{64 ( x + 4 )} - \frac{1}{256} ln ∣ x - 4 ∣ - \frac{1}{64 ( x - 4 )} + C

\frac{d}{d x} (x^{4} sin (2 x))

sin^{2} (e^{s i n^{2} (x)})

\frac{d}{d x} (\frac{2 x}{9})

\int e^{0.1 x} d x

n = 1 \sum \infty \frac{( 3 n + 2 )}{n}

What is the integral of 1/((16-x^2)^2) ?
The integral of 1/((16-x^2)^2) is 1/256 ln|x+4|-1/(64(x+4))-1/256 ln|x-4|-1/(64(x-4))+C