What is the integral of 4x^2-16x+16 ?

The integral of 4x^2-16x+16 is (4x^3)/3-8x^2+16x+C

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integral of 4x^2-16x+16

\int 4 x^{2} - 16 x + 16 d x

\frac{4 x ^{3}}{3} - 8 x^{2} + 16 x + C

Solution steps

\int 4 x^{2} - 16 x + 16 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 4 x^{2} d x - \int 16 x d x + \int 16 d x

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 16 x d x = 8 x^{2}

\int 16 d x = 16 x

= \frac{4 x ^{3}}{3} - 8 x^{2} + 16 x

Add a constant to the solution

= \frac{4 x ^{3}}{3} - 8 x^{2} + 16 x + C

x \to 1 lim (\frac{x ^{4} - 1}{x - 1})

f (- 2) = 2 x^{4}

\frac{d}{d x} ((4 x + 1)^{2})

\frac{d}{d x} (\frac{1}{( 1 - 2 x )})

e^{4 t}

What is the integral of 4x^2-16x+16 ?
The integral of 4x^2-16x+16 is (4x^3)/3-8x^2+16x+C