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

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

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

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

diverges

Solution steps

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

Factor

1 - x^{3} : - (- 1 + x^{3})

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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\int_{3}^{7} x e^{- 2 x} d x

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