What is the integral of (x-4)/((x-2)(x-3)) ?

The integral of (x-4)/((x-2)(x-3)) is 2ln|x-2|-ln|x-3|+C

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integral of (x-4)/((x-2)(x-3))

\int \frac{x - 4}{( x - 2 ) ( x - 3 )} d x

2 ln ∣ x - 2 ∣ - ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x - 4}{( x - 2 ) ( x - 3 )} d x

Take the partial fraction of

\frac{x - 4}{( x - 2 ) ( x - 3 )} : \frac{2}{x - 2} - \frac{1}{x - 3}

= \int \frac{2}{x - 2} - \frac{1}{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

= \int \frac{2}{x - 2} d x - \int \frac{1}{x - 3} d x

\int \frac{2}{x - 2} d x = 2 ln ∣ x - 2 ∣

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

= 2 ln ∣ x - 2 ∣ - ln ∣ x - 3 ∣

Add a constant to the solution

= 2 ln ∣ x - 2 ∣ - ln ∣ x - 3 ∣ + C

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What is the integral of (x-4)/((x-2)(x-3)) ?
The integral of (x-4)/((x-2)(x-3)) is 2ln|x-2|-ln|x-3|+C