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

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

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

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

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

Solution steps

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

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

\int \frac{3 x ^{3}}{x ^{2} + 1} d x

n = 0 \sum \infty \frac{1}{n}

\frac{\partial}{\partial x} (x^{0.7} y^{0.3})

\int_{0}^{1} 3 e^{- 2 x} d x

n = 1 \sum \infty \frac{7 \cdot n !}{n ^{n}}

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