What is the integral from 0 to 1 of (8y)/((x^3+1)^2) ?

The integral from 0 to 1 of (8y)/((x^3+1)^2) is 4/((1+x^3)^2)

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integral from 0 to 1 of (8y)/((x^3+1)^2)

\int_{0}^{1} \frac{8 y}{( x ^{3} + 1 ) ^{2}} d y

\frac{4}{( 1 + x ^{3} ) ^{2}}

Solution steps

\int_{0}^{1} \frac{8 y}{( x ^{3} + 1 ) ^{2}} d y

Take the constant out:

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

= \frac{8}{( 1 + x ^{3} ) ^{2}} \cdot \int_{0}^{1} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{0}^{1}

= \frac{8}{( 1 + x ^{3} ) ^{2}} [\frac{y ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= \frac{8}{( 1 + x ^{3} ) ^{2}} \cdot \frac{1}{2}

Simplify

= \frac{4}{( 1 + x ^{3} ) ^{2}}

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What is the integral from 0 to 1 of (8y)/((x^3+1)^2) ?
The integral from 0 to 1 of (8y)/((x^3+1)^2) is 4/((1+x^3)^2)