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

The integral of x/((x+1)(x+2)(x-3)) is 1/4 ln|x+1|-2/5 ln|x+2|+3/20 ln|x-3|+C

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

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

\frac{1}{4} ln ∣ x + 1 ∣ - \frac{2}{5} ln ∣ x + 2 ∣ + \frac{3}{20} ln ∣ x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x}{( x + 1 ) ( x + 2 ) ( x - 3 )} : \frac{1}{4 ( x + 1 )} - \frac{2}{5 ( x + 2 )} + \frac{3}{20 ( x - 3 )}

= \int \frac{1}{4 ( x + 1 )} - \frac{2}{5 ( x + 2 )} + \frac{3}{20 ( 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}{4 ( x + 1 )} d x - \int \frac{2}{5 ( x + 2 )} d x + \int \frac{3}{20 ( x - 3 )} d x

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

\int \frac{2}{5 ( x + 2 )} d x = \frac{2}{5} ln ∣ x + 2 ∣

\int \frac{3}{20 ( x - 3 )} d x = \frac{3}{20} ln ∣ x - 3 ∣

= \frac{1}{4} ln ∣ x + 1 ∣ - \frac{2}{5} ln ∣ x + 2 ∣ + \frac{3}{20} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ x + 1 ∣ - \frac{2}{5} ln ∣ x + 2 ∣ + \frac{3}{20} ln ∣ x - 3 ∣ + C

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What is the integral of x/((x+1)(x+2)(x-3)) ?
The integral of x/((x+1)(x+2)(x-3)) is 1/4 ln|x+1|-2/5 ln|x+2|+3/20 ln|x-3|+C