What is the integral of 1/(x^3+2x^2-5x-6) ?

The integral of 1/(x^3+2x^2-5x-6) is -1/6 ln|x+1|+1/15 ln|x-2|+1/10 ln|x+3|+C

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integral of 1/(x^3+2x^2-5x-6)

\int \frac{1}{x ^{3} + 2 x ^{2} - 5 x - 6} d x

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

Solution steps

\int \frac{1}{x ^{3} + 2 x ^{2} - 5 x - 6} d x

Take the partial fraction of

\frac{1}{x ^{3} + 2 x ^{2} - 5 x - 6} : - \frac{1}{6 ( x + 1 )} + \frac{1}{15 ( x - 2 )} + \frac{1}{10 ( x + 3 )}

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

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

\int \frac{1}{15 ( x - 2 )} d x = \frac{1}{15} ln ∣ x - 2 ∣

\int \frac{1}{10 ( x + 3 )} d x = \frac{1}{10} ln ∣ x + 3 ∣

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

Add a constant to the solution

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

What is the integral of 1/(x^3+2x^2-5x-6) ?
The integral of 1/(x^3+2x^2-5x-6) is -1/6 ln|x+1|+1/15 ln|x-2|+1/10 ln|x+3|+C