What is the integral of x/(x^3+8) ?

The integral of x/(x^3+8) is -1/6 ln|x+2|+1/12 (ln|x^2-2x+4|+2sqrt(3)arctan(1/(sqrt(3))(x-1)))+C

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integral of x/(x^3+8)

\int \frac{x}{x ^{3} + 8} d x

- \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{12} (ln x^{2} - 2 x + 4 + 23 arctan (\frac{1}{3} (x - 1))) + C

Solution steps

\int \frac{x}{x ^{3} + 8} d x

Take the partial fraction of

\frac{x}{x ^{3} + 8} : - \frac{1}{6 ( x + 2 )} + \frac{x + 2}{6 ( x ^{2} - 2 x + 4 )}

= \int - \frac{1}{6 ( x + 2 )} + \frac{x + 2}{6 ( x ^{2} - 2 x + 4 )} 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 + 2 )} d x + \int \frac{x + 2}{6 ( x ^{2} - 2 x + 4 )} d x

\int \frac{1}{6 ( x + 2 )} d x = \frac{1}{6} ln ∣ x + 2 ∣

\int \frac{x + 2}{6 ( x ^{2} - 2 x + 4 )} d x = \frac{1}{12} (ln x^{2} - 2 x + 4 + 23 arctan (\frac{1}{3} (x - 1)))

= - \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{12} (ln x^{2} - 2 x + 4 + 23 arctan (\frac{1}{3} (x - 1)))

Add a constant to the solution

= - \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{12} (ln x^{2} - 2 x + 4 + 23 arctan (\frac{1}{3} (x - 1))) + C

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What is the integral of x/(x^3+8) ?
The integral of x/(x^3+8) is -1/6 ln|x+2|+1/12 (ln|x^2-2x+4|+2sqrt(3)arctan(1/(sqrt(3))(x-1)))+C