What is the integral from 0 to 1 of (5x^2)/(2x^3+1) ?

The integral from 0 to 1 of (5x^2)/(2x^3+1) is 5/6 ln(3)

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integral from 0 to 1 of (5x^2)/(2x^3+1)

\int_{0}^{1} \frac{5 x ^{2}}{2 x ^{3} + 1} d x

\frac{5}{6} ln (3)

Decimal

0.91551 \dots

Solution steps

\int_{0}^{1} \frac{5 x ^{2}}{2 x ^{3} + 1} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} \frac{x ^{2}}{2 x ^{3} + 1} d x

Apply u-substitution

: \int_{1}^{3} \frac{1}{6 u} d u

= 5 \cdot \int_{1}^{3} \frac{1}{6 u} d u

Take the constant out:

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

= 5 \cdot \frac{1}{6} \cdot \int_{1}^{3} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{6} [ln ∣ u ∣]_{1}^{3}

Simplify

5 \cdot \frac{1}{6} [ln ∣ u ∣]_{1}^{3} : \frac{5}{6} [ln ∣ u ∣]_{1}^{3}

= \frac{5}{6} [ln ∣ u ∣]_{1}^{3}

Compute the boundaries:

ln (3)

= \frac{5}{6} ln (3)

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What is the integral from 0 to 1 of (5x^2)/(2x^3+1) ?
The integral from 0 to 1 of (5x^2)/(2x^3+1) is 5/6 ln(3)