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

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

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

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

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

Solution steps

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

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

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e t

\int_{0}^{1} x^{4} e^{3 x^{5}} d x

\frac{d}{d x} (\frac{4}{5 + cos ( x )})

f (x) = 8 (x - \frac{1}{x})^{2}, at x = 2

\frac{d}{d x} (1 0^{- 3 x^{2}})

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