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

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

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

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

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

Solution steps

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

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

(2 x^{2} - y e^{x}) d x - e^{x} d y = 0

\int \frac{ln ( t )}{t} d t

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

\frac{d}{d x} (x^{3} + 10 x^{2} + 25 x)

\int_{0}^{1} \frac{x}{4 - x ^{2}} d x

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