What is the integral from 0 to 1 of (x^3)/(1+x^4) ?

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

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integral from 0 to 1 of (x^3)/(1+x^4)

\int_{0}^{1} \frac{x ^{3}}{1 + x ^{4}} d x

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

Solution steps

\int_{0}^{1} \frac{x ^{3}}{1 + x ^{4}} d x

Apply u-substitution

: \int_{1}^{2} \frac{1}{4 u} d u

= \int_{1}^{2} \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_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

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

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