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

The integral from-1 to 2 of (x^2)/(x^3+2) is 1/3 ln(10)

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

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

\frac{1}{3} ln (10)

Decimal

0.76752 \dots

Solution steps

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

Apply u-substitution

= \int_{1}^{10} \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_{1}^{10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{1}^{10}

Compute the boundaries:

ln (10)

= \frac{1}{3} ln (10)

What is the integral from-1 to 2 of (x^2)/(x^3+2) ?
The integral from-1 to 2 of (x^2)/(x^3+2) is 1/3 ln(10)