What is the integral from 3 to 32 of 1/(x^3-27) ?

The integral from 3 to 32 of 1/(x^3-27) is diverges

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integral from 3 to 32 of 1/(x^3-27)

\int_{3}^{32} \frac{1}{x ^{3} - 27} d x

diverges

Solution steps

\int_{3}^{32} \frac{1}{x ^{3} - 27} d x

Take the partial fraction of

\frac{1}{x ^{3} - 27} : \frac{1}{27 ( x - 3 )} + \frac{- x - 6}{27 ( x ^{2} + 3 x + 9 )}

= \int_{3}^{32} \frac{1}{27 ( x - 3 )} + \frac{- x - 6}{27 ( x ^{2} + 3 x + 9 )} 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_{3}^{32} \frac{1}{27 ( x - 3 )} d x + \int_{3}^{32} \frac{- x - 6}{27 ( x ^{2} + 3 x + 9 )} d x

\int_{3}^{32} \frac{1}{27 ( x - 3 )} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

f (x) = \frac{5 x}{x + 3}, at x = 2

\frac{\partial}{\partial x} (ln (4 - t^{2}))

\int_{0}^{4} (10 - x) x d x

n = 0 \sum \infty \frac{2 ^{n}}{2 ^{n}}

e^{x} (x^{2} + 1) - 2 - x

What is the integral from 3 to 32 of 1/(x^3-27) ?
The integral from 3 to 32 of 1/(x^3-27) is diverges