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

\int \frac{x ^{3}}{1 + 9 x ^{2}} d x

\int e^{x} e^{2 x} d x

\int x ln (1 + \frac{1}{2 x}) d x

\int (x - 9) d x

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

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