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

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

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

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 0

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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\int_{- 1}^{3} 5 x x + 1 d x

\int_{0}^{x} (1 - \frac{r}{x})^{\frac{1}{n}} r d r

\int_{5}^{12} x d x

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