What is the integral from 0 to 1 of (x^2)/(x^3-1) ?

The integral from 0 to 1 of (x^2)/(x^3-1) is diverges

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integral from 0 to 1 of (x^2)/(x^3-1)

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

diverges

Solution steps

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

Apply u-substitution

: \int_{- 1}^{0} \frac{1}{3 u} d u

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

Take the constant out:

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

= \frac{1}{3} \cdot \int_{- 1}^{0} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{- 1}^{0}

Compute the boundaries:

diverges

= diverges

y = e^{5 x}

ln (cos (x))

x \to 2 lim (x^{2} - 5 x + 6)

(7 x^{2} - 12 x) e^{x}

\frac{d}{d x} (x^{3} (4 - x))

What is the integral from 0 to 1 of (x^2)/(x^3-1) ?
The integral from 0 to 1 of (x^2)/(x^3-1) is diverges