What is the integral from 0 to 4 of (x^2)/(x^3+8) ?

The integral from 0 to 4 of (x^2)/(x^3+8) is 2/3 ln(3)

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integral from 0 to 4 of (x^2)/(x^3+8)

\int_{0}^{4} \frac{x ^{2}}{x ^{3} + 8} d x

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

Decimal

0.73240 \dots

Solution steps

\int_{0}^{4} \frac{x ^{2}}{x ^{3} + 8} d x

Apply u-substitution

: \int_{8}^{72} \frac{1}{3 u} d u

= \int_{8}^{72} \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_{8}^{72} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{8}^{72}

Compute the boundaries:

ln (72) - 3 ln (2)

= \frac{1}{3} (ln (72) - 3 ln (2))

Simplify

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

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What is the integral from 0 to 4 of (x^2)/(x^3+8) ?
The integral from 0 to 4 of (x^2)/(x^3+8) is 2/3 ln(3)