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

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

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

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

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

Solution steps

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

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

\int_{0}^{\frac{11}{12}} 24 x d x

\frac{\partial}{\partial x} (x^{\frac{2}{3}} \cdot y^{\frac{1}{2}})

\frac{x ^{2} - 1}{x - 1}

- \frac{6}{( s - 2 ) ^{2}}

\frac{d}{d x} (e^{x y} + ln (\frac{x}{y}))

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