What is the integral from 2 to 13 of 1/(x^3-8) ?

The integral from 2 to 13 of 1/(x^3-8) is diverges

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integral from 2 to 13 of 1/(x^3-8)

\int_{2}^{13} \frac{1}{x ^{3} - 8} d x

diverges

Solution steps

\int_{2}^{13} \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_{2}^{13} \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_{2}^{13} \frac{1}{12 ( x - 2 )} d x + \int_{2}^{13} \frac{- x - 4}{12 ( x ^{2} + 2 x + 4 )} d x

\int_{2}^{13} \frac{1}{12 ( x - 2 )} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int (3 x + 4)^{\frac{1}{2}} d x

(x y^{2} + 3 y^{2}) d y - 2 x d x = 0

\frac{d}{d x} (\frac{5}{4 + x})

(4, 1), (7, 10)

\frac{d}{d x} (\frac{18}{x})

What is the integral from 2 to 13 of 1/(x^3-8) ?
The integral from 2 to 13 of 1/(x^3-8) is diverges