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

The integral of (x^2)/((x-1)(x+1)^2) is 1/4 ln|x-1|+3/4 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 (x^2)/((x-1)(x+1)^2)

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

\frac{1}{4} ln ∣ x - 1 ∣ + \frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + C

Solution steps

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

Take the partial fraction of

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

= \int \frac{1}{4 ( x - 1 )} + \frac{3}{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

= \int \frac{1}{4 ( x - 1 )} d x + \int \frac{3}{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{3}{4 ( x + 1 )} d x = \frac{3}{4} ln ∣ x + 1 ∣

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

= \frac{1}{4} ln ∣ x - 1 ∣ + \frac{3}{4} ln ∣ x + 1 ∣ - (- \frac{1}{2 ( x + 1 )})

Simplify

= \frac{1}{4} ln ∣ x - 1 ∣ + \frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )}

Add a constant to the solution

= \frac{1}{4} ln ∣ x - 1 ∣ + \frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + C

What is the integral of (x^2)/((x-1)(x+1)^2) ?
The integral of (x^2)/((x-1)(x+1)^2) is 1/4 ln|x-1|+3/4 ln|x+1|+1/(2(x+1))+C