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

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

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

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

diverges

Solution steps

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

Take the partial fraction of

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{2} e^{2 x} sin (3 x) d x

\int_{1}^{2} [\frac{I n x}{x ^{2}}] d x

\int_{0}^{1} \frac{1}{x + y} d y

\int_{\frac{π}{4}}^{\frac{π}{2}} - 2 x csc^{2} (x) d x

\int_{0}^{1} - ln (x) d x

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