What is the integral of (9x^2-8x+3) ?

The integral of (9x^2-8x+3) is 3x^3-4x^2+3x+C

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integral of (9x^2-8x+3)

\int (9 x^{2} - 8 x + 3) d x

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

Solution steps

\int 9 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 9 x^{2} d x - \int 8 x d x + \int 3 d x

\int 9 x^{2} d x = 3 x^{3}

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

\int cos (3 x) cos (2 x) d x

\int \frac{4}{x ^{2} - 4 x + 20} d x

x, x, 0, 2

\int (x^{\frac{3}{2}} + 5)^{6} x d x

\frac{d y}{d x} x^{2} = y - x y

What is the integral of (9x^2-8x+3) ?
The integral of (9x^2-8x+3) is 3x^3-4x^2+3x+C