What is the integral of (2x^3-4x+3x) ?

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

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integral of (2x^3-4x+3x)

\int (2 x^{3} - 4 x + 3 x) d x

\frac{x ^{4}}{2} - \frac{x ^{2}}{2} + C

Solution steps

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

Simplify

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

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

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

= \frac{x ^{4}}{2} - \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{4}}{2} - \frac{x ^{2}}{2} + C

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

\frac{d y}{d x} = \frac{e ^{- 2 x}}{1 + e ^{- 2 x}}

10 cos (x)

\int x e^{- \frac{1}{2} x^{2}} d x

x \to - \infty lim (e^{a x})

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