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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

= \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

= - \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 ∣

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

Add a constant to the solution

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

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