What is the integral from-2 to 2 of (3xy^2)/(x^2+1) ?

The integral from-2 to 2 of (3xy^2)/(x^2+1) is (16x)/(1+x^2)

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integral from-2 to 2 of (3xy^2)/(x^2+1)

\int_{- 2}^{2} \frac{3 x y ^{2}}{x ^{2} + 1} d y

\frac{16 x}{1 + x ^{2}}

Solution steps

\int_{- 2}^{2} \frac{3 x y ^{2}}{x ^{2} + 1} d y

Take the constant out:

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

= \frac{3 x}{1 + x ^{2}} \cdot \int_{- 2}^{2} y^{2} d y

Apply the Power Rule

: [\frac{y ^{3}}{3}]_{- 2}^{2}

= \frac{3 x}{1 + x ^{2}} [\frac{y ^{3}}{3}]_{- 2}^{2}

Compute the boundaries:

\frac{16}{3}

= \frac{3 x}{1 + x ^{2}} \cdot \frac{16}{3}

Simplify

= \frac{16 x}{1 + x ^{2}}

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What is the integral from-2 to 2 of (3xy^2)/(x^2+1) ?
The integral from-2 to 2 of (3xy^2)/(x^2+1) is (16x)/(1+x^2)