What is the integral of (6x^2-8x+3) ?

The integral of (6x^2-8x+3) is 2x^3-4x^2+3x+C

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integral of (6x^2-8x+3)

\int (6 x^{2} - 8 x + 3) d x

2 x^{3} - 4 x^{2} + 3 x + C

Solution steps

\int 6 x^{2} - 8 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 8 x d x + \int 3 d x

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

\int 8 x d x = 4 x^{2}

\int 3 d x = 3 x

= 2 x^{3} - 4 x^{2} + 3 x

Add a constant to the solution

= 2 x^{3} - 4 x^{2} + 3 x + C

\frac{d}{d x} (2 x tan (x) + x^{2} sec^{2} (x))

\frac{d}{d x} (cos (π x^{2}))

n = 1 \sum \infty \frac{1}{4 n ^{2} - 1}

- 2 x^{2} (x + 4)

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

What is the integral of (6x^2-8x+3) ?
The integral of (6x^2-8x+3) is 2x^3-4x^2+3x+C