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

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

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

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

\frac{1}{5} ln (2)

Decimal Notation

0.13862 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{5} [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= \frac{1}{5} ln (2)

\int_{- 2}^{0} (4 x^{3} + x^{2} - 2 x) d x

\int_{0}^{1} (2 t + 5) (e^{- t}) d t

\int_{0}^{\infty} π e^{- 2 x} d x

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

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

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