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

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

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

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

\frac{ln ( 2 ) y ^{2}}{2}

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Compute the boundaries:

ln (2)

= y^{2} \frac{1}{2} ln (2)

Simplify

= \frac{ln ( 2 ) y ^{2}}{2}

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