What is the integral of 1/((x-3)(x+2)(x-4)) ?

The integral of 1/((x-3)(x+2)(x-4)) is -1/5 ln|x-3|+1/30 ln|x+2|+1/6 ln|x-4|+C

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integral of 1/((x-3)(x+2)(x-4))

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

- \frac{1}{5} ln ∣ x - 3 ∣ + \frac{1}{30} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 4 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x - 3 ) ( x + 2 ) ( x - 4 )} : - \frac{1}{5 ( x - 3 )} + \frac{1}{30 ( x + 2 )} + \frac{1}{6 ( x - 4 )}

= \int - \frac{1}{5 ( x - 3 )} + \frac{1}{30 ( x + 2 )} + \frac{1}{6 ( 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}{5 ( x - 3 )} d x + \int \frac{1}{30 ( x + 2 )} d x + \int \frac{1}{6 ( x - 4 )} d x

\int \frac{1}{5 ( x - 3 )} d x = \frac{1}{5} ln ∣ x - 3 ∣

\int \frac{1}{30 ( x + 2 )} d x = \frac{1}{30} ln ∣ x + 2 ∣

\int \frac{1}{6 ( x - 4 )} d x = \frac{1}{6} ln ∣ x - 4 ∣

= - \frac{1}{5} ln ∣ x - 3 ∣ + \frac{1}{30} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{1}{5} ln ∣ x - 3 ∣ + \frac{1}{30} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 4 ∣ + C

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What is the integral of 1/((x-3)(x+2)(x-4)) ?
The integral of 1/((x-3)(x+2)(x-4)) is -1/5 ln|x-3|+1/30 ln|x+2|+1/6 ln|x-4|+C