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

The integral from 0 to 3 of (x^2)/(x^3+1) is 1/3 ln(28)

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

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

\frac{1}{3} ln (28)

Decimal

1.11073 \dots

Solution steps

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

Apply u-substitution

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

= \int_{1}^{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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{1}^{28}

Compute the boundaries:

ln (28)

= \frac{1}{3} ln (28)

\int_{0}^{1} cos (y^{2}) d y

\int_{2}^{\infty} \frac{x}{x ^{2} - 1} d x

\int_{0}^{\infty} \frac{1}{x - 1} d x

\int_{x^{2}}^{2 x} (x^{2} + y^{2}) d y

\int_{0}^{1} (x - 3)^{2} d x

What is the integral from 0 to 3 of (x^2)/(x^3+1) ?
The integral from 0 to 3 of (x^2)/(x^3+1) is 1/3 ln(28)