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

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

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

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

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

Decimal

0.50135 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{9} \frac{1}{3 u} d u

= \int_{2}^{9} \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_{2}^{9} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{2}^{9}

Compute the boundaries:

2 ln (3) - ln (2)

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

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