What is the integral of (5x-4)/(x^2-4x) ?

The integral of (5x-4)/(x^2-4x) is ln|x|+4ln|x-4|+C

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integral of (5x-4)/(x^2-4x)

\int \frac{5 x - 4}{x ^{2} - 4 x} d x

ln ∣ x ∣ + 4 ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{5 x - 4}{x ^{2} - 4 x} d x

Take the partial fraction of

\frac{5 x - 4}{x ^{2} - 4 x} : \frac{1}{x} + \frac{4}{x - 4}

= \int \frac{1}{x} + \frac{4}{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{1}{x} d x + \int \frac{4}{x - 4} d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ + 4 ln ∣ x - 4 ∣

Add a constant to the solution

= ln ∣ x ∣ + 4 ln ∣ x - 4 ∣ + C

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What is the integral of (5x-4)/(x^2-4x) ?
The integral of (5x-4)/(x^2-4x) is ln|x|+4ln|x-4|+C