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

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

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

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

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

Solution steps

\int \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 y d y

Apply the Power Rule

: \frac{y ^{2}}{2}

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

Simplify

\frac{8}{( 1 + x ^{3} ) ^{2}} \cdot \frac{y ^{2}}{2} : \frac{4 y ^{2}}{( 1 + x ^{3} ) ^{2}}

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

Add a constant to the solution

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

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