What is the integral from 0 to 5 of (x^3)/(x^4+4) ?

The integral from 0 to 5 of (x^3)/(x^4+4) is 1/4 (ln(629)-2ln(2))

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integral from 0 to 5 of (x^3)/(x^4+4)

\int_{0}^{5} \frac{x ^{3}}{x ^{4} + 4} d x

\frac{1}{4} (ln (629) - 2 ln (2))

Decimal Notation

1.26445 \dots

Solution steps

\int_{0}^{5} \frac{x ^{3}}{x ^{4} + 4} d x

Apply u-substitution

: \int_{4}^{629} \frac{1}{4 u} d u

= \int_{4}^{629} \frac{1}{4 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int_{4}^{629} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} [ln ∣ u ∣]_{4}^{629}

Compute the boundaries:

ln (629) - 2 ln (2)

= \frac{1}{4} (ln (629) - 2 ln (2))

\frac{d}{d x} ((x - 8)^{2})

3 x^{2}, \frac{1}{3} x, 0, 0.5

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4 sin (t)

\frac{d}{d x} (e^{3 x + 5})

What is the integral from 0 to 5 of (x^3)/(x^4+4) ?
The integral from 0 to 5 of (x^3)/(x^4+4) is 1/4 (ln(629)-2ln(2))