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

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

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

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

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

Solution steps

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

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

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