What is the integral of 4\sqrt[3]{16-4x}-8 ?

The integral of 4\sqrt[3]{16-4x}-8 is -3/4 (16-4x)^{4/3}-8x+C

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integral of 4\sqrt[3]{16-4x}-8

- \frac{3}{4} (16 - 4 x)^{\frac{4}{3}} - 8 x + C

Solution steps

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int 8 d x = 8 x

= - \frac{3}{4} (16 - 4 x)^{\frac{4}{3}} - 8 x

Add a constant to the solution

= - \frac{3}{4} (16 - 4 x)^{\frac{4}{3}} - 8 x + C

\frac{\partial}{\partial x} (sin (x) \cdot cos (x))

x \to + (- 1) lim (x + 2 x^{4})

- \frac{1}{( x - 5 ) ^{2}}

\frac{d}{d x} (e^{x^{0}})

\int_{1}^{3} \frac{x}{( 2 x - 1 ) ^{3}} d x

What is the integral of 4\sqrt[3]{16-4x}-8 ?
The integral of 4\sqrt[3]{16-4x}-8 is -3/4 (16-4x)^{4/3}-8x+C