What is the integral of 1/(x^3-8) ?

The integral of 1/(x^3-8) is 1/12 ln|x-2|-1/24 ln|x^2+2x+4|-1/(4sqrt(3))arctan(1/(sqrt(3))(x+1))+C

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integral of 1/(x^3-8)

\int \frac{1}{x ^{3} - 8} d x

\frac{1}{12} ln ∣ x - 2 ∣ - \frac{1}{24} ln x^{2} + 2 x + 4 - \frac{1}{4 3} arctan (\frac{1}{3} (x + 1)) + C

Solution steps

\int \frac{1}{x ^{3} - 8} d x

Take the partial fraction of

\frac{1}{x ^{3} - 8} : \frac{1}{12 ( x - 2 )} + \frac{- x - 4}{12 ( x ^{2} + 2 x + 4 )}

= \int \frac{1}{12 ( x - 2 )} + \frac{- x - 4}{12 ( 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}{12 ( x - 2 )} d x + \int \frac{- x - 4}{12 ( x ^{2} + 2 x + 4 )} d x

\int \frac{1}{12 ( x - 2 )} d x = \frac{1}{12} ln ∣ x - 2 ∣

\int \frac{- x - 4}{12 ( x ^{2} + 2 x + 4 )} d x = \frac{1}{24 3} (- 3 ln x^{2} + 2 x + 4 - 6 arctan (\frac{1}{3} (x + 1)))

= \frac{1}{12} ln ∣ x - 2 ∣ + \frac{1}{24 3} (- 3 ln x^{2} + 2 x + 4 - 6 arctan (\frac{1}{3} (x + 1)))

Expand

\frac{1}{24 3} (- 3 ln x^{2} + 2 x + 4 - 6 arctan (\frac{1}{3} (x + 1))) : - \frac{1}{24} ln x^{2} + 2 x + 4 - \frac{1}{4 3} arctan (\frac{1}{3} (x + 1))

= \frac{1}{12} ln ∣ x - 2 ∣ - \frac{1}{24} ln x^{2} + 2 x + 4 - \frac{1}{4 3} arctan (\frac{1}{3} (x + 1))

Add a constant to the solution

= \frac{1}{12} ln ∣ x - 2 ∣ - \frac{1}{24} ln x^{2} + 2 x + 4 - \frac{1}{4 3} arctan (\frac{1}{3} (x + 1)) + C

f (x) = (7 x^{2} - 11 x) e^{x}

\frac{\partial}{\partial z} (x - z - arctan (yz))

\int 6 x sec (x) tan (x) d x

\frac{d}{d x} (60 x^{4} - 180 x^{3} + 180 x^{2} - 60 x)

\frac{d}{d x} (ln (x^{2} + 10))

What is the integral of 1/(x^3-8) ?
The integral of 1/(x^3-8) is 1/12 ln|x-2|-1/24 ln|x^2+2x+4|-1/(4sqrt(3))arctan(1/(sqrt(3))(x+1))+C