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

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

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

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

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

Decimal Notation

0.73240 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{9} \frac{1}{3 u} d u

= \int_{1}^{9} \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}^{9} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{1}^{9}

Compute the boundaries:

2 ln (3)

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

Simplify

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

\int_{0}^{4} \frac{1}{9 - 2 x} d x

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

\int_{- 2}^{2} 4 - y^{2} d y

\int_{0}^{1} - x^{2} + 1 d x

\int_{- 1}^{1} \frac{1}{5 - 4 x} d x

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