What is the integral from 1 to 2 of (x^2)/(x^3-8) ?

The integral from 1 to 2 of (x^2)/(x^3-8) is diverges

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integral from 1 to 2 of (x^2)/(x^3-8)

\int_{1}^{2} \frac{x ^{2}}{x ^{3} - 8} d x

diverges

Solution steps

\int_{1}^{2} \frac{x ^{2}}{x ^{3} - 8} d x

Apply u-substitution

: \int_{- 7}^{0} \frac{1}{3 u} d u

= \int_{- 7}^{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_{- 7}^{0} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{- 7}^{0}

Compute the boundaries:

diverges

= diverges

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

\int_{2}^{8} 4 x d x

What is the integral from 1 to 2 of (x^2)/(x^3-8) ?
The integral from 1 to 2 of (x^2)/(x^3-8) is diverges