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

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

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

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

diverges

Solution steps

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

Compute the indefinite integral:

\int \frac{1}{x ^{3} - 1} d x = \frac{1}{3} ln ∣ x - 1 ∣ - \frac{1}{6} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 x + 1)) + C

Compute the boundaries:

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

diverges

= diverges

\int y x + y^{2} d x

\int (3 sec (x) - cos (x))^{2} d x

x^{2} - 4 x + 39

\int_{1}^{4} π (\frac{1}{x})^{2} d x

y^{^{'}} = 3 x^{2} y

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