What is the integral of-4x^2-12x+16 ?

The integral of-4x^2-12x+16 is -(4x^3)/3-6x^2+16x+C

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integral of-4x^2-12x+16

\int - 4 x^{2} - 12 x + 16 d x

- \frac{4 x ^{3}}{3} - 6 x^{2} + 16 x + C

Solution steps

\int - 4 x^{2} - 12 x + 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 4 x^{2} d x - \int 12 x d x + \int 16 d x

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 12 x d x = 6 x^{2}

\int 16 d x = 16 x

= - \frac{4 x ^{3}}{3} - 6 x^{2} + 16 x

Add a constant to the solution

= - \frac{4 x ^{3}}{3} - 6 x^{2} + 16 x + C

\frac{d}{d x} (\frac{0.14 x}{0.015 + x})

\int y - e^{c o s (x)} sin (x) d x

\int \frac{( 3 x - 5 )}{x ^{3} - 5 x ^{2} - 14 x} d x

\frac{d}{d x} (3 x^{3} + y^{3})

(x cos (x))^{^{'}}

What is the integral of-4x^2-12x+16 ?
The integral of-4x^2-12x+16 is -(4x^3)/3-6x^2+16x+C