What is the integral from 2 to 4 of (x^2)/(x^3-1) ?

The integral from 2 to 4 of (x^2)/(x^3-1) is 2/3 ln(3)

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integral from 2 to 4 of (x^2)/(x^3-1)

\int_{2}^{4} \frac{x ^{2}}{x ^{3} - 1} d x

\frac{2}{3} ln (3)

Decimal

0.73240 \dots

Solution steps

\int_{2}^{4} \frac{x ^{2}}{x ^{3} - 1} d x

Apply u-substitution

: \int_{7}^{63} \frac{1}{3 u} d u

= \int_{7}^{63} \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}^{63} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{7}^{63}

Compute the boundaries:

ln (63) - ln (7)

= \frac{1}{3} (ln (63) - ln (7))

Simplify

= \frac{2}{3} ln (3)

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What is the integral from 2 to 4 of (x^2)/(x^3-1) ?
The integral from 2 to 4 of (x^2)/(x^3-1) is 2/3 ln(3)