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

The integral from 0 to 2 of (y^3)/(y^4+1) is 1/4 ln(17)

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integral from 0 to 2 of (y^3)/(y^4+1)

\int_{0}^{2} \frac{y ^{3}}{y ^{4} + 1} d y

\frac{1}{4} ln (17)

Decimal Notation

0.70830 \dots

Solution steps

\int_{0}^{2} \frac{y ^{3}}{y ^{4} + 1} d y

Apply u-substitution

: \int_{1}^{17} \frac{1}{4 u} d u

= \int_{1}^{17} \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}^{17} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} [ln ∣ u ∣]_{1}^{17}

Compute the boundaries:

ln (17)

= \frac{1}{4} ln (17)

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What is the integral from 0 to 2 of (y^3)/(y^4+1) ?
The integral from 0 to 2 of (y^3)/(y^4+1) is 1/4 ln(17)