What is the integral of 8/((x-9)(x-1)) ?

The integral of 8/((x-9)(x-1)) is ln|x-9|-ln|x-1|+C

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integral of 8/((x-9)(x-1))

\int \frac{8}{( x - 9 ) ( x - 1 )} d x

ln ∣ x - 9 ∣ - ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{8}{( x - 9 ) ( x - 1 )} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{( x - 9 ) ( x - 1 )} d x

Take the partial fraction of

\frac{1}{( x - 9 ) ( x - 1 )} : \frac{1}{8 ( x - 9 )} - \frac{1}{8 ( x - 1 )}

= 8 \cdot \int \frac{1}{8 ( x - 9 )} - \frac{1}{8 ( x - 1 )} 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

= 8 (\int \frac{1}{8 ( x - 9 )} d x - \int \frac{1}{8 ( x - 1 )} d x)

\int \frac{1}{8 ( x - 9 )} d x = \frac{1}{8} ln ∣ x - 9 ∣

\int \frac{1}{8 ( x - 1 )} d x = \frac{1}{8} ln ∣ x - 1 ∣

= 8 (\frac{1}{8} ln ∣ x - 9 ∣ - \frac{1}{8} ln ∣ x - 1 ∣)

Simplify

8 (\frac{1}{8} ln ∣ x - 9 ∣ - \frac{1}{8} ln ∣ x - 1 ∣) : ln ∣ x - 9 ∣ - ln ∣ x - 1 ∣

= ln ∣ x - 9 ∣ - ln ∣ x - 1 ∣

Add a constant to the solution

= ln ∣ x - 9 ∣ - ln ∣ x - 1 ∣ + C

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What is the integral of 8/((x-9)(x-1)) ?
The integral of 8/((x-9)(x-1)) is ln|x-9|-ln|x-1|+C