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

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

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

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

ln ∣ x - 6 ∣ - ln ∣ x - 4 ∣ + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

= ln ∣ x - 6 ∣ - ln ∣ x - 4 ∣

Add a constant to the solution

= ln ∣ x - 6 ∣ - ln ∣ x - 4 ∣ + C

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