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

The integral of (3x-x^2+2/3 x^3) is (3x^2)/2-(x^3)/3+(x^4)/6+C

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integral of (3x-x^2+2/3 x^3)

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

\frac{3 x ^{2}}{2} - \frac{x ^{3}}{3} + \frac{x ^{4}}{6} + C

Solution steps

\int 3 x - x^{2} + \frac{2}{3} x^{3} 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 x^{2} d x + \int \frac{2}{3} x^{3} d x

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

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

\int \frac{2}{3} x^{3} d x = \frac{x ^{4}}{6}

= \frac{3 x ^{2}}{2} - \frac{x ^{3}}{3} + \frac{x ^{4}}{6}

Add a constant to the solution

= \frac{3 x ^{2}}{2} - \frac{x ^{3}}{3} + \frac{x ^{4}}{6} + C

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What is the integral of (3x-x^2+2/3 x^3) ?
The integral of (3x-x^2+2/3 x^3) is (3x^2)/2-(x^3)/3+(x^4)/6+C