What is the integral from 0 to 1 of (6x^2)/(1-x^3) ?

The integral from 0 to 1 of (6x^2)/(1-x^3) is diverges

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integral from 0 to 1 of (6x^2)/(1-x^3)

\int_{0}^{1} \frac{6 x ^{2}}{1 - x ^{3}} d x

diverges

Solution steps

\int_{0}^{1} \frac{6 x ^{2}}{1 - x ^{3}} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} \frac{x ^{2}}{1 - x ^{3}} d x

Apply u-substitution

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

= 6 \cdot \int_{1}^{0} - \frac{1}{3 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 6 (- \int_{0}^{1} - \frac{1}{3 u} d u)

Take the constant out:

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

= 6 (- (- \frac{1}{3} \cdot \int_{0}^{1} \frac{1}{u} d u))

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 (- (- \frac{1}{3} [ln ∣ u ∣]_{0}^{1}))

Simplify

6 (- (- \frac{1}{3} [ln ∣ u ∣]_{0}^{1})) : 2 [ln ∣ u ∣]_{0}^{1}

= 2 [ln ∣ u ∣]_{0}^{1}

Compute the boundaries:

diverges

= diverges

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What is the integral from 0 to 1 of (6x^2)/(1-x^3) ?
The integral from 0 to 1 of (6x^2)/(1-x^3) is diverges