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

The integral of 1/(x(x+2)(x-3)) is -1/6 ln|x|+1/10 ln|x+2|+1/15 ln|x-3|+C

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

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

- \frac{1}{6} ln ∣ x ∣ + \frac{1}{10} ln ∣ x + 2 ∣ + \frac{1}{15} ln ∣ x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{x ( x + 2 ) ( x - 3 )} : - \frac{1}{6 x} + \frac{1}{10 ( x + 2 )} + \frac{1}{15 ( x - 3 )}

= \int - \frac{1}{6 x} + \frac{1}{10 ( x + 2 )} + \frac{1}{15 ( x - 3 )} 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}{6 x} d x + \int \frac{1}{10 ( x + 2 )} d x + \int \frac{1}{15 ( x - 3 )} d x

\int \frac{1}{6 x} d x = \frac{1}{6} ln ∣ x ∣

\int \frac{1}{10 ( x + 2 )} d x = \frac{1}{10} ln ∣ x + 2 ∣

\int \frac{1}{15 ( x - 3 )} d x = \frac{1}{15} ln ∣ x - 3 ∣

= - \frac{1}{6} ln ∣ x ∣ + \frac{1}{10} ln ∣ x + 2 ∣ + \frac{1}{15} ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ x ∣ + \frac{1}{10} ln ∣ x + 2 ∣ + \frac{1}{15} ln ∣ x - 3 ∣ + C

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\frac{\partial}{\partial x} (a x^{b} y^{1 - b})

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What is the integral of 1/(x(x+2)(x-3)) ?
The integral of 1/(x(x+2)(x-3)) is -1/6 ln|x|+1/10 ln|x+2|+1/15 ln|x-3|+C