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

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

g (t) = t^{2} t

\frac{d}{d x} (2^{x} (π x^{2} + cos (x)))

x \to 9 - lim (\frac{x - 9}{∣ x - 9 ∣})

\frac{d}{d x} (\frac{x}{x ^{- 1} + 7})

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