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

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

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

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

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

Solution steps

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

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

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

\int 2 d x = 2 x

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

Add a constant to the solution

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

x \to 0 lim (x^{2} \cdot e^{\frac{1}{x}})

\frac{d}{d x} ((x^{- 3} + 4) (x^{- 2} - 5 x))

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

\int 1 + y^{2} d y

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