What is the integral of 4x^3-3x^2+6x-1 ?

The integral of 4x^3-3x^2+6x-1 is x^4-x^3+3x^2-x+C

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integral of 4x^3-3x^2+6x-1

\int 4 x^{3} - 3 x^{2} + 6 x - 1 d x

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

Solution steps

\int 4 x^{3} - 3 x^{2} + 6 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^{3} d x - \int 3 x^{2} d x + \int 6 x d x - \int 1 d x

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

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

\int 6 x d x = 3 x^{2}

\int 1 d x = x

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

Add a constant to the solution

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

e^{x} - e^{- x}

\frac{d}{d x} (e^{5 x y})

\int 14 (x - 5)^{6} d x

\frac{\partial}{\partial θ} (z)

\frac{d}{d x} ((2 x - 5)^{- 1} \cdot (4 x + 3)^{- 1})

What is the integral of 4x^3-3x^2+6x-1 ?
The integral of 4x^3-3x^2+6x-1 is x^4-x^3+3x^2-x+C