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

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

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integral of (x^2-3x+4)

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

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

Solution steps

\int x^{2} - 3 x + 4 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 4 d x

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

\int - tan (x) sin (x) d x

4 x y^{^{'}} - 16 y = x^{- 7}, y (1) = 7

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

y = \frac{8}{sin ( x )} + \frac{1}{cot ( x )}

x \to 0 lim (\frac{5}{1 + e ^{\frac{1}{x}}})

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