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

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

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

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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