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

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

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

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

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

Solution steps

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

\int 2 d x = 2 x

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

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

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

Add a constant to the solution

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

f (x) = - 4 - 3 x^{2}, (2, - 16)

\int_{- 45}^{0} \frac{1}{4 - x} d x

n = 1 \sum \infty 8^{- n}

\frac{\partial}{\partial x} (1 + e^{- x y} cos (y))

\int 2 cos (4 x) cos (2 x) d x

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