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

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

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

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

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

Solution steps

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

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

\frac{d}{d t} (\frac{2 t - 1}{2 t + 1})

\frac{2}{s ^{5}}

x \to 0 lim (sin (x^{12}))

\frac{d}{d x} (x arcsec (x^{3}))

\frac{( 6 + 6 s )}{2 s ^{2} + 1}

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