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

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

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

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

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

Solution steps

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

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

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