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

f (x) = (x^{2} + x + 1)^{3}

sin (ω \cdot t)

\int \frac{x ^{2} + x}{x - 1} d x

\frac{\partial}{\partial x} (5 - y^{3} - x^{7})

\frac{d y}{d x} = y^{2} - 9, y (1) = 4