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

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

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

Take the partial fraction of

\frac{x ^{4}}{x ^{4} - 1} : 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

\frac{d}{d t} (e^{- 3 t} sin (3 t))

y = x^{\frac{7}{5}} + e^{- 4 x}

\frac{d}{d t} (t + 1)

\frac{d}{d x} (4 \frac{1}{x})

(- 4.3) (3.1)

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