What is the integral from 0 to 3 of (x^2)/(x^3+9) ?

The integral from 0 to 3 of (x^2)/(x^3+9) is 2/3 ln(2)

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integral from 0 to 3 of (x^2)/(x^3+9)

\int_{0}^{3} \frac{x ^{2}}{x ^{3} + 9} d x

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

Decimal Notation

0.46209 \dots

Solution steps

\int_{0}^{3} \frac{x ^{2}}{x ^{3} + 9} d x

Apply u-substitution

: \int_{9}^{36} \frac{1}{3 u} d u

= \int_{9}^{36} \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_{9}^{36} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{9}^{36}

Compute the boundaries:

2 ln (6) - 2 ln (3)

= \frac{1}{3} (2 ln (6) - 2 ln (3))

Simplify

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

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

\int_{0}^{x} \frac{x ^{2}}{9} d x

\int_{1}^{e} \frac{9}{x} d x

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

\int_{0}^{7} \frac{3}{x + 9} d x

What is the integral from 0 to 3 of (x^2)/(x^3+9) ?
The integral from 0 to 3 of (x^2)/(x^3+9) is 2/3 ln(2)