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

The integral from 0 to 2 of (x^2)/(x^3+8) is 1/3 ln(2)

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

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

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

Decimal

0.23104 \dots

Solution steps

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

Apply u-substitution

: \int_{8}^{16} \frac{1}{3 u} d u

= \int_{8}^{16} \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}^{16} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{8}^{16}

Compute the boundaries:

ln (2)

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

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\int_{0}^{1} - 2 x + 1 d x

\int_{- 1}^{4} 3 x^{2} d x

\int_{2}^{3} 2 e^{- 2 x} d x

What is the integral from 0 to 2 of (x^2)/(x^3+8) ?
The integral from 0 to 2 of (x^2)/(x^3+8) is 1/3 ln(2)