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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Take the constant out:

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

= - 2 \cdot \int \frac{x ^{2}}{x ^{4} - 1} d x

Take the partial fraction of

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

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

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

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

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

Simplify

- 2 (\frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) : - arctan (x) + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣

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

Add a constant to the solution

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

\int (8 x^{3} y - y^{3}) d x

\int \frac{12 x ^{3} - 3 x ^{2}}{3 x ^{4} - x ^{3}} d x

\int \frac{2 x ^{4} - x - 4}{x ^{3} + 8} d x

\int \frac{e ^{12 t}}{4 e ^{6 t} - 1} d t

\int \frac{6 x ^{2} - 1}{x} d x

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