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

x \to 0 lim (sin (x) ln (2 x))

\frac{1}{1 + u ^{2}} d u = d x

\frac{d}{d x} (\frac{7 x}{6})

\frac{\partial}{\partial x} (6 e^{x^{5} y^{3}})

\int 10 x^{3} y d y

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