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

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

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

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

\frac{1}{3} (ln (e^{3} + e) - 1)

Decimal Notation

0.70897 \dots

Solution steps

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

Apply u-substitution

: \int_{e}^{e^{3} + e} \frac{1}{3 u} d u

= \int_{e}^{e^{3} + e} \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_{e}^{e^{3} + e} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{e}^{e^{3} + e}

Compute the boundaries:

ln (e^{3} + e) - 1

= \frac{1}{3} (ln (e^{3} + e) - 1)

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