What is the integral of 1/(x^2(x^2-9)) ?

The integral of 1/(x^2(x^2-9)) is 1/(9x)-1/54 ln|x+3|+1/54 ln|x-3|+C

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integral of 1/(x^2(x^2-9))

\int \frac{1}{x ^{2} ( x ^{2} - 9 )} d x

\frac{1}{9 x} - \frac{1}{54} ln ∣ x + 3 ∣ + \frac{1}{54} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{1}{x ^{2} ( x ^{2} - 9 )} d x

Take the partial fraction of

\frac{1}{x ^{2} ( x ^{2} - 9 )} : - \frac{1}{9 x ^{2}} - \frac{1}{54 ( x + 3 )} + \frac{1}{54 ( x - 3 )}

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

\int \frac{1}{9 x ^{2}} d x = - \frac{1}{9 x}

\int \frac{1}{54 ( x + 3 )} d x = \frac{1}{54} ln ∣ x + 3 ∣

\int \frac{1}{54 ( x - 3 )} d x = \frac{1}{54} ln ∣ x - 3 ∣

= - (- \frac{1}{9 x}) - \frac{1}{54} ln ∣ x + 3 ∣ + \frac{1}{54} ln ∣ x - 3 ∣

Simplify

= \frac{1}{9 x} - \frac{1}{54} ln ∣ x + 3 ∣ + \frac{1}{54} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{1}{9 x} - \frac{1}{54} ln ∣ x + 3 ∣ + \frac{1}{54} ln ∣ x - 3 ∣ + C

What is the integral of 1/(x^2(x^2-9)) ?
The integral of 1/(x^2(x^2-9)) is 1/(9x)-1/54 ln|x+3|+1/54 ln|x-3|+C