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

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

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

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

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

Decimal

0.23104 \dots

Solution steps

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

\int_{0}^{3} x e^{2 x} d x

\int_{0}^{4} \frac{1}{( x - 3 ) ^{2}} d x

\int_{0}^{2} - x^{2} + 2 x d x

\int_{0}^{0.3} sin (x^{2}) d x

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

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