What is the integral from 0 to 1 of (5x^4)/(x^5+6) ?

The integral from 0 to 1 of (5x^4)/(x^5+6) is ln(7)-ln(6)

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integral from 0 to 1 of (5x^4)/(x^5+6)

\int_{0}^{1} \frac{5 x ^{4}}{x ^{5} + 6} d x

ln (7) - ln (6)

Decimal Notation

0.15415 \dots

Solution steps

\int_{0}^{1} \frac{5 x ^{4}}{x ^{5} + 6} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} \frac{x ^{4}}{x ^{5} + 6} d x

Apply u-substitution

: \int_{6}^{7} \frac{1}{5 u} d u

= 5 \cdot \int_{6}^{7} \frac{1}{5 u} d u

Take the constant out:

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

= 5 \cdot \frac{1}{5} \cdot \int_{6}^{7} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{5} [ln ∣ u ∣]_{6}^{7}

Simplify

5 \cdot \frac{1}{5} [ln ∣ u ∣]_{6}^{7} : [ln ∣ u ∣]_{6}^{7}

= [ln ∣ u ∣]_{6}^{7}

Compute the boundaries:

ln (7) - ln (6)

= ln (7) - ln (6)

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

\int_{0}^{\frac{π}{2}} 3 x cos (2 x) d x

What is the integral from 0 to 1 of (5x^4)/(x^5+6) ?
The integral from 0 to 1 of (5x^4)/(x^5+6) is ln(7)-ln(6)