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

The integral from 2 to 3 of (x^2)/(x^3-2) is 1/3 (2ln(5)-ln(6))

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

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

\frac{1}{3} (2 ln (5) - ln (6))

Decimal

0.47570 \dots

Solution steps

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

Apply u-substitution

= \int_{6}^{25} \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_{6}^{25} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{6}^{25}

Compute the boundaries:

2 ln (5) - ln (6)

= \frac{1}{3} (2 ln (5) - ln (6))

What is the integral from 2 to 3 of (x^2)/(x^3-2) ?
The integral from 2 to 3 of (x^2)/(x^3-2) is 1/3 (2ln(5)-ln(6))