What is the integral of (x-12)/(x^2-4x) ?

The integral of (x-12)/(x^2-4x) is 3ln|x|-2ln|x-4|+C

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integral of (x-12)/(x^2-4x)

\int \frac{x - 12}{x ^{2} - 4 x} d x

3 ln ∣ x ∣ - 2 ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{x - 12}{x ^{2} - 4 x} d x

Take the partial fraction of

\frac{x - 12}{x ^{2} - 4 x} : \frac{3}{x} - \frac{2}{x - 4}

= \int \frac{3}{x} - \frac{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

= \int \frac{3}{x} d x - \int \frac{2}{x - 4} d x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

= 3 ln ∣ x ∣ - 2 ln ∣ x - 4 ∣

Add a constant to the solution

= 3 ln ∣ x ∣ - 2 ln ∣ x - 4 ∣ + C

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What is the integral of (x-12)/(x^2-4x) ?
The integral of (x-12)/(x^2-4x) is 3ln|x|-2ln|x-4|+C