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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Take the partial fraction of

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

= \int - 1 + \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

= - \int 1 d x + \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 1 d x = 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 ∣

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

Add a constant to the solution

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

\int sec (\frac{θ}{2}) tan (\frac{θ}{2}) d θ

\int \frac{6}{( 2 - x ) x ^{2} - 4 x + 3} d x

\int \frac{( 3 x + 1 )}{x ^{2}} d x

\int \frac{8}{x ^{2} x ^{2} - 16} d x

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

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