What is the integral of (3x^2-3y^2) ?

The integral of (3x^2-3y^2) is 3x^2y-y^3+C

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integral of (3x^2-3y^2)

\int (3 x^{2} - 3 y^{2}) d y

3 x^{2} y - y^{3} + C

Solution steps

\int 3 x^{2} - 3 y^{2} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 3 x^{2} d y - \int 3 y^{2} d y

\int 3 x^{2} d y = 3 x^{2} y

\int 3 y^{2} d y = y^{3}

= 3 x^{2} y - y^{3}

Add a constant to the solution

= 3 x^{2} y - y^{3} + C

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What is the integral of (3x^2-3y^2) ?
The integral of (3x^2-3y^2) is 3x^2y-y^3+C