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

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

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

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

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

Solution steps

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

Expand

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

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

\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 ∣

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

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

Simplify

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

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

Add a constant to the solution

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

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

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

\frac{d}{d x} (e^{- x} cos (y))

\frac{d}{d x} ((1 + x^{3})^{5})

\frac{\partial}{\partial y} (e^{y} sin (x) y)

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