What is the integral from 0 to 1 of (x^4)/(2(x^5+1)) ?

The integral from 0 to 1 of (x^4)/(2(x^5+1)) is 1/10 ln(2)

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integral from 0 to 1 of (x^4)/(2(x^5+1))

\int_{0}^{1} \frac{x ^{4}}{2 ( x ^{5} + 1 )} d x

\frac{1}{10} ln (2)

Decimal

0.06931 \dots

Solution steps

\int_{0}^{1} \frac{x ^{4}}{2 ( x ^{5} + 1 )} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} \frac{x ^{4}}{x ^{5} + 1} d x

Apply u-substitution

: \int_{1}^{2} \frac{1}{5 u} d u

= \frac{1}{2} \cdot \int_{1}^{2} \frac{1}{5 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{5} \cdot \int_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} \cdot \frac{1}{5} [ln ∣ u ∣]_{1}^{2}

Simplify

\frac{1}{2} \cdot \frac{1}{5} [ln ∣ u ∣]_{1}^{2} : \frac{1}{10} [ln ∣ u ∣]_{1}^{2}

= \frac{1}{10} [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= \frac{1}{10} ln (2)

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What is the integral from 0 to 1 of (x^4)/(2(x^5+1)) ?
The integral from 0 to 1 of (x^4)/(2(x^5+1)) is 1/10 ln(2)