What is the integral of (6x^2+4x-3) ?

The integral of (6x^2+4x-3) is 2x^3+2x^2-3x+C

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integral of (6x^2+4x-3)

\int (6 x^{2} + 4 x - 3) d x

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

Solution steps

\int 6 x^{2} + 4 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 4 x d x - \int 3 d x

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

\int 4 x d x = 2 x^{2}

\int 3 d x = 3 x

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

Add a constant to the solution

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

What is the integral of (6x^2+4x-3) ?
The integral of (6x^2+4x-3) is 2x^3+2x^2-3x+C