What is the integral from 0 to 2 of 1/(x^3-x) ?

The integral from 0 to 2 of 1/(x^3-x) is diverges

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integral from 0 to 2 of 1/(x^3-x)

\int_{0}^{2} \frac{1}{x ^{3} - x} d x

diverges

Solution steps

\int_{0}^{2} \frac{1}{x ^{3} - x} d x

Find undefined (singularity) points:

x = 1

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{1} \frac{1}{x ^{3} - x} d x + \int_{1}^{2} \frac{1}{x ^{3} - x} d x

\int_{0}^{1} \frac{1}{x ^{3} - x} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{6}^{22} 2 f (\frac{x}{2}) d x

\int_{2}^{0} x d x

\int_{4}^{6} \frac{1}{x ^{2} - 4} d x

\int_{0}^{7} π x d x

\int_{0}^{2} \frac{x ^{3}}{x ^{4} + 1} d x

What is the integral from 0 to 2 of 1/(x^3-x) ?
The integral from 0 to 2 of 1/(x^3-x) is diverges