What is the integral of 4/((x-6)(x-2)) ?

The integral of 4/((x-6)(x-2)) is ln|x-6|-ln|x-2|+C

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integral of 4/((x-6)(x-2))

\int \frac{4}{( x - 6 ) ( x - 2 )} d x

ln ∣ x - 6 ∣ - ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{4}{( x - 6 ) ( x - 2 )} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{( x - 6 ) ( x - 2 )} d x

Take the partial fraction of

\frac{1}{( x - 6 ) ( x - 2 )} : \frac{1}{4 ( x - 6 )} - \frac{1}{4 ( x - 2 )}

= 4 \cdot \int \frac{1}{4 ( x - 6 )} - \frac{1}{4 ( x - 2 )} 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

= 4 (\int \frac{1}{4 ( x - 6 )} d x - \int \frac{1}{4 ( x - 2 )} d x)

\int \frac{1}{4 ( x - 6 )} d x = \frac{1}{4} ln ∣ x - 6 ∣

\int \frac{1}{4 ( x - 2 )} d x = \frac{1}{4} ln ∣ x - 2 ∣

= 4 (\frac{1}{4} ln ∣ x - 6 ∣ - \frac{1}{4} ln ∣ x - 2 ∣)

Simplify

4 (\frac{1}{4} ln ∣ x - 6 ∣ - \frac{1}{4} ln ∣ x - 2 ∣) : ln ∣ x - 6 ∣ - ln ∣ x - 2 ∣

= ln ∣ x - 6 ∣ - ln ∣ x - 2 ∣

Add a constant to the solution

= ln ∣ x - 6 ∣ - ln ∣ x - 2 ∣ + C

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What is the integral of 4/((x-6)(x-2)) ?
The integral of 4/((x-6)(x-2)) is ln|x-6|-ln|x-2|+C