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

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

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

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

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

Solution steps

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

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

\frac{d}{d x} (cos (2 - 3 x))

\frac{d y}{d t} + e^{t} y = e^{t}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 2 y = 0

x \to - \frac{7 π}{2} - lim (sec (x))

\frac{d y}{d x} = 9 - \frac{y}{6 x + 250}

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