What is the integral from 4 to 69 of 1/(x^3-64) ?

The integral from 4 to 69 of 1/(x^3-64) is diverges

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integral from 4 to 69 of 1/(x^3-64)

\int_{4}^{69} \frac{1}{x ^{3} - 64} d x

diverges

Solution steps

\int_{4}^{69} \frac{1}{x ^{3} - 64} d x

Take the partial fraction of

\frac{1}{x ^{3} - 64} : \frac{1}{48 ( x - 4 )} + \frac{- x - 8}{48 ( x ^{2} + 4 x + 16 )}

= \int_{4}^{69} \frac{1}{48 ( x - 4 )} + \frac{- x - 8}{48 ( x ^{2} + 4 x + 16 )} 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_{4}^{69} \frac{1}{48 ( x - 4 )} d x + \int_{4}^{69} \frac{- x - 8}{48 ( x ^{2} + 4 x + 16 )} d x

\int_{4}^{69} \frac{1}{48 ( x - 4 )} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

x \to 0 lim (x^{2} + 4)

f (x) = e^{x} (3 x^{2} - 6 x + 6)

y^{^{'}} = (t + y)^{2} - 1, y (1) = 5

\frac{d}{d x} (\frac{2}{x ^{2} - 4})

\frac{d}{d x} (e^{x} arcsin (x))

What is the integral from 4 to 69 of 1/(x^3-64) ?
The integral from 4 to 69 of 1/(x^3-64) is diverges