What is the integral of-4x^3-8x^2-4 ?

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

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integral of-4x^3-8x^2-4

\int - 4 x^{3} - 8 x^{2} - 4 d x

- x^{4} - \frac{8 x ^{3}}{3} - 4 x + C

Solution steps

\int - 4 x^{3} - 8 x^{2} - 4 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^{3} d x - \int 8 x^{2} d x - \int 4 d x

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

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

\int 4 d x = 4 x

= - x^{4} - \frac{8 x ^{3}}{3} - 4 x

Add a constant to the solution

= - x^{4} - \frac{8 x ^{3}}{3} - 4 x + C

\frac{d}{d x} (- x^{2} + 5 x)

\frac{\partial ^{2}}{\partial x ^{2}} (x)

x \to 3 - lim (\frac{x}{x - 3})

\int_{1}^{\infty} 3 x^{- 4} d x

\int_{0}^{1} x e^{6 x} d x

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