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

\frac{\partial}{\partial x} (ln (x^{10} y))

f (x) = x^{2} - 2 x + 2

\frac{1 - x}{2}

(x 1 + y^{2}) d x - x^{2} d y = 0

\int \frac{4}{( x ^{2} ) ( x ^{2} + 4 )} d x

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