What is the integral of (-9x^2-4x+1) ?

The integral of (-9x^2-4x+1) is -3x^3-2x^2+x+C

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integral of (-9x^2-4x+1)

\int (- 9 x^{2} - 4 x + 1) d x

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

Solution steps

\int - 9 x^{2} - 4 x + 1 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 4 x d x + \int 1 d x

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\int (cos (5 x) - sec^{2} (x)) d x

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

\int x - 3 x d x

What is the integral of (-9x^2-4x+1) ?
The integral of (-9x^2-4x+1) is -3x^3-2x^2+x+C