What is the integral of 1/((x^2-1)(x^2-4)) ?

The integral of 1/((x^2-1)(x^2-4)) is 1/6 ln|x+1|-1/6 ln|x-1|-1/12 ln|x+2|+1/12 ln|x-2|+C

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integral of 1/((x^2-1)(x^2-4))

\int \frac{1}{( x ^{2} - 1 ) ( x ^{2} - 4 )} d x

\frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{1}{( x ^{2} - 1 ) ( x ^{2} - 4 )} d x

Take the partial fraction of

\frac{1}{( x ^{2} - 1 ) ( x ^{2} - 4 )} : \frac{1}{6 ( x + 1 )} - \frac{1}{6 ( x - 1 )} - \frac{1}{12 ( x + 2 )} + \frac{1}{12 ( x - 2 )}

= \int \frac{1}{6 ( x + 1 )} - \frac{1}{6 ( x - 1 )} - \frac{1}{12 ( x + 2 )} + \frac{1}{12 ( x - 2 )} 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}{6 ( x + 1 )} d x - \int \frac{1}{6 ( x - 1 )} d x - \int \frac{1}{12 ( x + 2 )} d x + \int \frac{1}{12 ( x - 2 )} d x

\int \frac{1}{6 ( x + 1 )} d x = \frac{1}{6} ln ∣ x + 1 ∣

\int \frac{1}{6 ( x - 1 )} d x = \frac{1}{6} ln ∣ x - 1 ∣

\int \frac{1}{12 ( x + 2 )} d x = \frac{1}{12} ln ∣ x + 2 ∣

\int \frac{1}{12 ( x - 2 )} d x = \frac{1}{12} ln ∣ x - 2 ∣

= \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣ + C

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What is the integral of 1/((x^2-1)(x^2-4)) ?
The integral of 1/((x^2-1)(x^2-4)) is 1/6 ln|x+1|-1/6 ln|x-1|-1/12 ln|x+2|+1/12 ln|x-2|+C