What is the integral of (4x^2-8x+1) ?

The integral of (4x^2-8x+1) is (4x^3)/3-4x^2+x+C

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integral of (4x^2-8x+1)

\int (4 x^{2} - 8 x + 1) d x

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

Solution steps

\int 4 x^{2} - 8 x + 1 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 8 x d x + \int 1 d x

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

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

\int 1 d x = x

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

Add a constant to the solution

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

n = 0 \sum \infty (1 - \frac{1}{n})^{n}

\frac{d x}{d t} = \frac{3}{x}

x \to 0 - lim ((\frac{1}{2})^{\frac{1}{x}})

x \to - 3 lim (8 x^{2})

\int \frac{1}{9 x ^{3} - 2 x ^{2}} d x

What is the integral of (4x^2-8x+1) ?
The integral of (4x^2-8x+1) is (4x^3)/3-4x^2+x+C