What is the integral from 0 to 4 of (x^2)/(x^3+4) ?

The integral from 0 to 4 of (x^2)/(x^3+4) is 1/3 ln(17)

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integral from 0 to 4 of (x^2)/(x^3+4)

\int_{0}^{4} \frac{x ^{2}}{x ^{3} + 4} d x

\frac{1}{3} ln (17)

Decimal

0.94440 \dots

Solution steps

\int_{0}^{4} \frac{x ^{2}}{x ^{3} + 4} d x

Apply u-substitution

: \int_{4}^{68} \frac{1}{3 u} d u

= \int_{4}^{68} \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_{4}^{68} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{4}^{68}

Compute the boundaries:

ln (68) - 2 ln (2)

= \frac{1}{3} (ln (68) - 2 ln (2))

Simplify

= \frac{1}{3} ln (17)

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

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\int_{1}^{3} \frac{1}{x - 5} d x

\int_{1}^{8} \frac{9}{x} d x

What is the integral from 0 to 4 of (x^2)/(x^3+4) ?
The integral from 0 to 4 of (x^2)/(x^3+4) is 1/3 ln(17)