What is the integral of (x^3-3x^2+6x-20) ?

The integral of (x^3-3x^2+6x-20) is (x^4)/4-x^3+3x^2-20x+C

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integral of (x^3-3x^2+6x-20)

\int (x^{3} - 3 x^{2} + 6 x - 20) d x

\frac{x ^{4}}{4} - x^{3} + 3 x^{2} - 20 x + C

Solution steps

\int x^{3} - 3 x^{2} + 6 x - 20 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 3 x^{2} d x + \int 6 x d x - \int 20 d x

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

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

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

\int 20 d x = 20 x

= \frac{x ^{4}}{4} - x^{3} + 3 x^{2} - 20 x

Add a constant to the solution

= \frac{x ^{4}}{4} - x^{3} + 3 x^{2} - 20 x + C

What is the integral of (x^3-3x^2+6x-20) ?
The integral of (x^3-3x^2+6x-20) is (x^4)/4-x^3+3x^2-20x+C