What is the integral of (x^2+7x+2)/((x+1)(x-2)(x+3)) ?

The integral of (x^2+7x+2)/((x+1)(x-2)(x+3)) is 2/3 ln|x+1|+4/3 ln|x-2|-ln|x+3|+C

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integral of (x^2+7x+2)/((x+1)(x-2)(x+3))

\int \frac{x ^{2} + 7 x + 2}{( x + 1 ) ( x - 2 ) ( x + 3 )} d x

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

Solution steps

\int \frac{x ^{2} + 7 x + 2}{( x + 1 ) ( x - 2 ) ( x + 3 )} d x

Take the partial fraction of

\frac{x ^{2} + 7 x + 2}{( x + 1 ) ( x - 2 ) ( x + 3 )} : \frac{2}{3 ( x + 1 )} + \frac{4}{3 ( x - 2 )} - \frac{1}{x + 3}

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

\int \frac{2}{3 ( x + 1 )} d x = \frac{2}{3} ln ∣ x + 1 ∣

\int \frac{4}{3 ( x - 2 )} d x = \frac{4}{3} ln ∣ x - 2 ∣

\int \frac{1}{x + 3} d x = ln ∣ x + 3 ∣

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

Add a constant to the solution

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

What is the integral of (x^2+7x+2)/((x+1)(x-2)(x+3)) ?
The integral of (x^2+7x+2)/((x+1)(x-2)(x+3)) is 2/3 ln|x+1|+4/3 ln|x-2|-ln|x+3|+C