What is the integral of-2x+9x^2-4x^3+16 ?

The integral of-2x+9x^2-4x^3+16 is -x^2+3x^3-x^4+16x+C

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integral of-2x+9x^2-4x^3+16

\int - 2 x + 9 x^{2} - 4 x^{3} + 16 d x

- x^{2} + 3 x^{3} - x^{4} + 16 x + C

Solution steps

\int - 2 x + 9 x^{2} - 4 x^{3} + 16 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 2 x d x + \int 9 x^{2} d x - \int 4 x^{3} d x + \int 16 d x

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

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

\int 4 x^{3} d x = x^{4}

\int 16 d x = 16 x

= - x^{2} + 3 x^{3} - x^{4} + 16 x

Add a constant to the solution

= - x^{2} + 3 x^{3} - x^{4} + 16 x + C

\int e^{7 x} sin (3 x) d x

\int 8 tan^{3} (x) sec (x) d x

\frac{d}{d x} (3 x^{2} - 10)

x^{2} y^{^{''}} - 7 x y^{^{'}} + 15 y = 0, y (1) = 0, y^{^{'}} (1) = 2

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

What is the integral of-2x+9x^2-4x^3+16 ?
The integral of-2x+9x^2-4x^3+16 is -x^2+3x^3-x^4+16x+C