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

The integral of (12x+18)/((x+2)(x+4)(x-1)) is ln|x+2|-3ln|x+4|+2ln|x-1|+C

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integral of (12x+18)/((x+2)(x+4)(x-1))

\int \frac{12 x + 18}{( x + 2 ) ( x + 4 ) ( x - 1 )} d x

ln ∣ x + 2 ∣ - 3 ln ∣ x + 4 ∣ + 2 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{12 x + 18}{( x + 2 ) ( x + 4 ) ( x - 1 )} d x

Take the partial fraction of

\frac{12 x + 18}{( x + 2 ) ( x + 4 ) ( x - 1 )} : \frac{1}{x + 2} - \frac{3}{x + 4} + \frac{2}{x - 1}

= \int \frac{1}{x + 2} - \frac{3}{x + 4} + \frac{2}{x - 1} 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 + 2} d x - \int \frac{3}{x + 4} d x + \int \frac{2}{x - 1} d x

\int \frac{1}{x + 2} d x = ln ∣ x + 2 ∣

\int \frac{3}{x + 4} d x = 3 ln ∣ x + 4 ∣

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

= ln ∣ x + 2 ∣ - 3 ln ∣ x + 4 ∣ + 2 ln ∣ x - 1 ∣

Add a constant to the solution

= ln ∣ x + 2 ∣ - 3 ln ∣ x + 4 ∣ + 2 ln ∣ x - 1 ∣ + C

What is the integral of (12x+18)/((x+2)(x+4)(x-1)) ?
The integral of (12x+18)/((x+2)(x+4)(x-1)) is ln|x+2|-3ln|x+4|+2ln|x-1|+C