What is the integral from 0 to 4 of (x^2)/(x^3+6) ?

The integral from 0 to 4 of (x^2)/(x^3+6) is 1/3 ln(35/3)

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integral from 0 to 4 of (x^2)/(x^3+6)

\int_{0}^{4} \frac{x ^{2}}{x ^{3} + 6} d x

\frac{1}{3} ln (\frac{35}{3})

Decimal

0.81891 \dots

Solution steps

\int_{0}^{4} \frac{x ^{2}}{x ^{3} + 6} d x

Apply u-substitution

= \int_{6}^{70} \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_{6}^{70} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{6}^{70}

Compute the boundaries:

ln (70) - ln (6)

= \frac{1}{3} (ln (70) - ln (6))

Simplify

= \frac{1}{3} ln (\frac{35}{3})

What is the integral from 0 to 4 of (x^2)/(x^3+6) ?
The integral from 0 to 4 of (x^2)/(x^3+6) is 1/3 ln(35/3)