What is the integral of (9x^2-8x+4) ?

The integral of (9x^2-8x+4) is 3x^3-4x^2+4x+C

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integral of (9x^2-8x+4)

\int (9 x^{2} - 8 x + 4) d x

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

Solution steps

\int 9 x^{2} - 8 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 9 x^{2} d x - \int 8 x d x + \int 4 d x

\int 9 x^{2} d x = 3 x^{3}

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

n = 1 \sum \infty 8 (\frac{1}{5})^{n}

\frac{d}{d x} ((5 x)^{7 x})

x \to 0 + lim (\frac{5 x}{sin ( x )})

\frac{d}{d t} (e^{r t})

\int e^{- 4 l n (x)} d x

What is the integral of (9x^2-8x+4) ?
The integral of (9x^2-8x+4) is 3x^3-4x^2+4x+C