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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{3}{10 ( x - 2 )} d x = \frac{3}{10} ln ∣ x - 2 ∣

\int \frac{2}{15 ( x + 3 )} d x = \frac{2}{15} ln ∣ x + 3 ∣

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

Add a constant to the solution

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

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