What is the integral from 0 to 1 of (x^2)/(x^3+7) ?

The integral from 0 to 1 of (x^2)/(x^3+7) is ln(2)-1/3 ln(7)

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integral from 0 to 1 of (x^2)/(x^3+7)

\int_{0}^{1} \frac{x ^{2}}{x ^{3} + 7} d x

ln (2) - \frac{1}{3} ln (7)

Decimal Notation

0.04451 \dots

Solution steps

\int_{0}^{1} \frac{x ^{2}}{x ^{3} + 7} d x

Apply u-substitution

: \int_{7}^{8} \frac{1}{3 u} d u

= \int_{7}^{8} \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}^{8} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{7}^{8}

Compute the boundaries:

3 ln (2) - ln (7)

= \frac{1}{3} (3 ln (2) - ln (7))

Simplify

= ln (2) - \frac{1}{3} ln (7)

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What is the integral from 0 to 1 of (x^2)/(x^3+7) ?
The integral from 0 to 1 of (x^2)/(x^3+7) is ln(2)-1/3 ln(7)