What is the integral of x^2-8x+12 ?

The integral of x^2-8x+12 is (x^3)/3-4x^2+12x+C

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integral of x^2-8x+12

\int x^{2} - 8 x + 12 d x

\frac{x ^{3}}{3} - 4 x^{2} + 12 x + C

Solution steps

\int x^{2} - 8 x + 12 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 8 x d x + \int 12 d x

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

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

\int 12 d x = 12 x

= \frac{x ^{3}}{3} - 4 x^{2} + 12 x

Add a constant to the solution

= \frac{x ^{3}}{3} - 4 x^{2} + 12 x + C

\frac{d}{d x} (ln (x + 1 - x^{2}))

n = 0 \sum \infty (\frac{5}{4}) (\frac{2}{3})^{n}

n = 0 \sum \infty \frac{n}{n ^{2} + 2}

\int (\frac{x ^{2}}{2} - \frac{2}{x ^{2}}) d x

\frac{\partial}{\partial x} (\frac{3 y}{x ^{2} + 2})

What is the integral of x^2-8x+12 ?
The integral of x^2-8x+12 is (x^3)/3-4x^2+12x+C