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

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

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integral of 1/((1-x^3))

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

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

Solution steps

\int \frac{1}{1 - x ^{3}} d x

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

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