What is the integral of (8x)/((x-1)(x+1)^2) ?

The integral of (8x)/((x-1)(x+1)^2) is 8(1/4 ln|x-1|-1/4 ln|x+1|-1/(2(x+1)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (8x)/((x-1)(x+1)^2)

\int \frac{8 x}{( x - 1 ) ( x + 1 ) ^{2}} d x

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

Solution steps

\int \frac{8 x}{( x - 1 ) ( x + 1 ) ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{x}{( x - 1 ) ( x + 1 ) ^{2}} d x

Take the partial fraction of

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

= 8 \cdot \int \frac{1}{4 ( x - 1 )} - \frac{1}{4 ( x + 1 )} + \frac{1}{2 ( x + 1 ) ^{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

= 8 (\int \frac{1}{4 ( x - 1 )} d x - \int \frac{1}{4 ( x + 1 )} d x + \int \frac{1}{2 ( x + 1 ) ^{2}} d x)

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

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

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

= 8 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )})

Add a constant to the solution

= 8 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )}) + C

What is the integral of (8x)/((x-1)(x+1)^2) ?
The integral of (8x)/((x-1)(x+1)^2) is 8(1/4 ln|x-1|-1/4 ln|x+1|-1/(2(x+1)))+C