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

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

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

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

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

Decimal

0.23104 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

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

x \to 0.01 lim (\frac{e ^{x} - 1}{x})

\frac{d}{d x} (\frac{sin ( \frac{1}{x} ) x ^{3}}{ln ( x )})

1 - x^{2}, x^{6} - x + 1

x \to - \infty lim (x^{4} + x)

sin (2 x) d x + cos (8 y) d y = 0

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