What is the integral of 12-(16-2\sqrt[3]{96-8x}) ?

The integral of 12-(16-2\sqrt[3]{96-8x}) is -4x-3(-x+12)^{4/3}+C

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integral of 12-(16-2\sqrt[3]{96-8x})

- 4 x - 3 (- x + 12)^{\frac{4}{3}} + 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 12 d x = 12 x

= 12 x - (16 x + 3 (- x + 12)^{\frac{4}{3}})

Simplify

= - 4 x - 3 (- x + 12)^{\frac{4}{3}}

Add a constant to the solution

= - 4 x - 3 (- x + 12)^{\frac{4}{3}} + C

36 y^{^{''}} + 36 y^{^{'}} + 11 y = 0, y (0) = 6, y^{^{'}} (0) = 5

\frac{6}{3 x + 1}, at a = - 1

\int (cos (x) + \frac{1}{8} x) d x

x \to 3 - lim (e^{\frac{2}{3 - x}})

\frac{\partial}{\partial y} (x^{y} + y^{x})

What is the integral of 12-(16-2\sqrt[3]{96-8x}) ?
The integral of 12-(16-2\sqrt[3]{96-8x}) is -4x-3(-x+12)^{4/3}+C