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

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

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integral of (3x^2-4x+1)

\int (3 x^{2} - 4 x + 1) d x

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

Solution steps

\int 3 x^{2} - 4 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 3 x^{2} d x - \int 4 x d x + \int 1 d x

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\frac{d}{d y} (2 k q y (y^{2} + d^{2})^{- \frac{3}{2}})

\int π^{4} x^{2} d x

\frac{d}{d x} (\frac{3 x ^{4}}{e ^{x}})

3 y^{2} = 2 y (x^{4} - x) \frac{d y}{d x} - 3

\int_{2}^{3} \frac{30}{3 - x} d x

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