What is the integral of (18-12x)/((4x-1)(x-4)) ?

The integral of (18-12x)/((4x-1)(x-4)) is -ln|4x-1|-2ln|x-4|+C

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integral of (18-12x)/((4x-1)(x-4))

\int \frac{18 - 12 x}{( 4 x - 1 ) ( x - 4 )} d x

- ln ∣ 4 x - 1 ∣ - 2 ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{18 - 12 x}{( 4 x - 1 ) ( x - 4 )} d x

Take the partial fraction of

\frac{18 - 12 x}{( 4 x - 1 ) ( x - 4 )} : - \frac{4}{4 x - 1} - \frac{2}{x - 4}

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

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

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

= - ln ∣ 4 x - 1 ∣ - 2 ln ∣ x - 4 ∣

Add a constant to the solution

= - ln ∣ 4 x - 1 ∣ - 2 ln ∣ x - 4 ∣ + C

What is the integral of (18-12x)/((4x-1)(x-4)) ?
The integral of (18-12x)/((4x-1)(x-4)) is -ln|4x-1|-2ln|x-4|+C