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

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

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integral of x^2y^2-3x

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

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

Solution steps

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

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 d x + \int y^{2} x^{2} d x

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

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

= - \frac{3 x ^{2}}{2} + \frac{y ^{2} x ^{3}}{3}

Add a constant to the solution

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

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