What is the integral of 1/(x^4-81) ?

The integral of 1/(x^4-81) is -1/54 arctan(x/3)-1/108 ln|x+3|+1/108 ln|x-3|+C

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integral of 1/(x^4-81)

\int \frac{1}{x ^{4} - 81} d x

- \frac{1}{54} arctan (\frac{x}{3}) - \frac{1}{108} ln ∣ x + 3 ∣ + \frac{1}{108} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{1}{x ^{4} - 81} d x

Take the partial fraction of

\frac{1}{x ^{4} - 81} : - \frac{1}{18 ( x ^{2} + 9 )} - \frac{1}{108 ( x + 3 )} + \frac{1}{108 ( x - 3 )}

= \int - \frac{1}{18 ( x ^{2} + 9 )} - \frac{1}{108 ( x + 3 )} + \frac{1}{108 ( x - 3 )} 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}{18 ( x ^{2} + 9 )} d x - \int \frac{1}{108 ( x + 3 )} d x + \int \frac{1}{108 ( x - 3 )} d x

\int \frac{1}{18 ( x ^{2} + 9 )} d x = \frac{1}{54} arctan (\frac{x}{3})

\int \frac{1}{108 ( x + 3 )} d x = \frac{1}{108} ln ∣ x + 3 ∣

\int \frac{1}{108 ( x - 3 )} d x = \frac{1}{108} ln ∣ x - 3 ∣

= - \frac{1}{54} arctan (\frac{x}{3}) - \frac{1}{108} ln ∣ x + 3 ∣ + \frac{1}{108} ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{1}{54} arctan (\frac{x}{3}) - \frac{1}{108} ln ∣ x + 3 ∣ + \frac{1}{108} ln ∣ x - 3 ∣ + C

What is the integral of 1/(x^4-81) ?
The integral of 1/(x^4-81) is -1/54 arctan(x/3)-1/108 ln|x+3|+1/108 ln|x-3|+C