What is the integral of x/(x^3+3x^2-9x-27) ?

The integral of x/(x^3+3x^2-9x-27) is -1/12 ln|x+3|-1/(2(x+3))+1/12 ln|x-3|+C

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integral of x/(x^3+3x^2-9x-27)

\int \frac{x}{x ^{3} + 3 x ^{2} - 9 x - 27} d x

- \frac{1}{12} ln ∣ x + 3 ∣ - \frac{1}{2 ( x + 3 )} + \frac{1}{12} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x}{x ^{3} + 3 x ^{2} - 9 x - 27} d x

Take the partial fraction of

\frac{x}{x ^{3} + 3 x ^{2} - 9 x - 27} : - \frac{1}{12 ( x + 3 )} + \frac{1}{2 ( x + 3 ) ^{2}} + \frac{1}{12 ( x - 3 )}

= \int - \frac{1}{12 ( x + 3 )} + \frac{1}{2 ( x + 3 ) ^{2}} + \frac{1}{12 ( 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}{12 ( x + 3 )} d x + \int \frac{1}{2 ( x + 3 ) ^{2}} d x + \int \frac{1}{12 ( x - 3 )} d x

\int \frac{1}{12 ( x + 3 )} d x = \frac{1}{12} ln ∣ x + 3 ∣

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

\int \frac{1}{12 ( x - 3 )} d x = \frac{1}{12} ln ∣ x - 3 ∣

= - \frac{1}{12} ln ∣ x + 3 ∣ - \frac{1}{2 ( x + 3 )} + \frac{1}{12} ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{1}{12} ln ∣ x + 3 ∣ - \frac{1}{2 ( x + 3 )} + \frac{1}{12} ln ∣ x - 3 ∣ + C

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What is the integral of x/(x^3+3x^2-9x-27) ?
The integral of x/(x^3+3x^2-9x-27) is -1/12 ln|x+3|-1/(2(x+3))+1/12 ln|x-3|+C