What is the integral of x^3-6x^2+8x ?

The integral of x^3-6x^2+8x is (x^4)/4-2x^3+4x^2+C

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integral of x^3-6x^2+8x

\int x^{3} - 6 x^{2} + 8 x d x

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

Solution steps

\int x^{3} - 6 x^{2} + 8 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 x^{3} d x - \int 6 x^{2} d x + \int 8 x d x

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

\int 6 x^{2} d x = 2 x^{3}

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

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

Add a constant to the solution

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

x \to 0 lim (e^{3 x} - 1)

\int 4 x e^{- 2 x} d x

(x^{2} - 25 ln (x^{2}))^{^{'}}

\int \frac{10}{15 sin ( x ) - 8 cos ( x )} d x

What is the integral of x^3-6x^2+8x ?
The integral of x^3-6x^2+8x is (x^4)/4-2x^3+4x^2+C