What is the integral from 0 to 4 of (x^2)/(x^3+7) ?

The integral from 0 to 4 of (x^2)/(x^3+7) is 1/3 (ln(71)-ln(7))

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integral from 0 to 4 of (x^2)/(x^3+7)

\int_{0}^{4} \frac{x ^{2}}{x ^{3} + 7} d x

\frac{1}{3} (ln (71) - ln (7))

Decimal Notation

0.77225 \dots

Solution steps

\int_{0}^{4} \frac{x ^{2}}{x ^{3} + 7} d x

Apply u-substitution

: \int_{7}^{71} \frac{1}{3 u} d u

= \int_{7}^{71} \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_{7}^{71} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{7}^{71}

Compute the boundaries:

ln (71) - ln (7)

= \frac{1}{3} (ln (71) - ln (7))

f (x) = \frac{( e ^{x} )}{( 1 - e ^{x} )}

\int x^{2} (ln (x)) d x

\int (\frac{d}{1} (y))^{2} d x

\frac{( 4 s - 5 )}{s ^{2} - 6 s + 13}

n = 1 \sum \infty (\frac{1}{3})^{n}

What is the integral from 0 to 4 of (x^2)/(x^3+7) ?
The integral from 0 to 4 of (x^2)/(x^3+7) is 1/3 (ln(71)-ln(7))