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

The integral from 1 to 3 of (3x^2)/(x^3+1) is ln(14)

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

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

ln (14)

Decimal Notation

2.63905 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

: \int_{2}^{28} \frac{1}{3 u} d u

= 3 \cdot \int_{2}^{28} \frac{1}{3 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \frac{1}{3} \cdot \int_{2}^{28} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{3} [ln ∣ u ∣]_{2}^{28}

Simplify

3 \cdot \frac{1}{3} [ln ∣ u ∣]_{2}^{28} : [ln ∣ u ∣]_{2}^{28}

= [ln ∣ u ∣]_{2}^{28}

Compute the boundaries:

ln (28) - ln (2)

= ln (28) - ln (2)

Simplify

= ln (14)

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What is the integral from 1 to 3 of (3x^2)/(x^3+1) ?
The integral from 1 to 3 of (3x^2)/(x^3+1) is ln(14)