What is the integral of 6/((x-9)(x-3)) ?

The integral of 6/((x-9)(x-3)) is ln|x-9|-ln|x-3|+C

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integral of 6/((x-9)(x-3))

\int \frac{6}{( x - 9 ) ( x - 3 )} d x

ln ∣ x - 9 ∣ - ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{6}{( x - 9 ) ( x - 3 )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{1}{( x - 9 ) ( x - 3 )} d x

Take the partial fraction of

\frac{1}{( x - 9 ) ( x - 3 )} : \frac{1}{6 ( x - 9 )} - \frac{1}{6 ( x - 3 )}

= 6 \cdot \int \frac{1}{6 ( x - 9 )} - \frac{1}{6 ( 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

= 6 (\int \frac{1}{6 ( x - 9 )} d x - \int \frac{1}{6 ( x - 3 )} d x)

\int \frac{1}{6 ( x - 9 )} d x = \frac{1}{6} ln ∣ x - 9 ∣

\int \frac{1}{6 ( x - 3 )} d x = \frac{1}{6} ln ∣ x - 3 ∣

= 6 (\frac{1}{6} ln ∣ x - 9 ∣ - \frac{1}{6} ln ∣ x - 3 ∣)

Simplify

6 (\frac{1}{6} ln ∣ x - 9 ∣ - \frac{1}{6} ln ∣ x - 3 ∣) : ln ∣ x - 9 ∣ - ln ∣ x - 3 ∣

= ln ∣ x - 9 ∣ - ln ∣ x - 3 ∣

Add a constant to the solution

= ln ∣ x - 9 ∣ - ln ∣ x - 3 ∣ + C

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What is the integral of 6/((x-9)(x-3)) ?
The integral of 6/((x-9)(x-3)) is ln|x-9|-ln|x-3|+C