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

y^{^{''}} + 3 y^{^{'}} + 2 y = 15 + e x

\int x - x^{3} d x

\frac{d}{d x} (a c e^{- e^{c (b - x)} + c (b - x)})

\int tan (x) sec^{6} (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