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

The integral from 0 to 2 of (x^3)/(x^4+6) is 1/4 ln(11/3)

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

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

\frac{1}{4} ln (\frac{11}{3})

Decimal Notation

0.32482 \dots

Solution steps

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

Apply u-substitution

: \int_{6}^{22} \frac{1}{4 u} d u

= \int_{6}^{22} \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{6}^{22} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} [ln ∣ u ∣]_{6}^{22}

Compute the boundaries:

ln (22) - ln (6)

= \frac{1}{4} (ln (22) - ln (6))

Simplify

= \frac{1}{4} ln (\frac{11}{3})

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What is the integral from 0 to 2 of (x^3)/(x^4+6) ?
The integral from 0 to 2 of (x^3)/(x^4+6) is 1/4 ln(11/3)