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

The integral of 1/((x-9)(x+9)) is 1/18 ln|x-9|-1/18 ln|x+9|+C

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integral of 1/((x-9)(x+9))

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

\frac{1}{18} ln ∣ x - 9 ∣ - \frac{1}{18} ln ∣ x + 9 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x - 9 ) ( x + 9 )} : \frac{1}{18 ( x - 9 )} - \frac{1}{18 ( x + 9 )}

= \int \frac{1}{18 ( x - 9 )} - \frac{1}{18 ( x + 9 )} 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 - 9 )} d x - \int \frac{1}{18 ( x + 9 )} d x

\int \frac{1}{18 ( x - 9 )} d x = \frac{1}{18} ln ∣ x - 9 ∣

\int \frac{1}{18 ( x + 9 )} d x = \frac{1}{18} ln ∣ x + 9 ∣

= \frac{1}{18} ln ∣ x - 9 ∣ - \frac{1}{18} ln ∣ x + 9 ∣

Add a constant to the solution

= \frac{1}{18} ln ∣ x - 9 ∣ - \frac{1}{18} ln ∣ x + 9 ∣ + C

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What is the integral of 1/((x-9)(x+9)) ?
The integral of 1/((x-9)(x+9)) is 1/18 ln|x-9|-1/18 ln|x+9|+C