What is the integral of (3x^2-6x+3) ?

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

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integral of (3x^2-6x+3)

\int (3 x^{2} - 6 x + 3) d x

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

Solution steps

\int 3 x^{2} - 6 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 6 x d x + \int 3 d x

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

\int_{0}^{\frac{1}{4}} 5 arcsin (4 x) d x

\int \frac{6 x ^{3} + 5 x ^{2} - 47 x - 36}{x ^{2} - 9} d x

10 - x, 1

arccot (\frac{x}{6})

x \to - \infty lim (x^{4})

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