What is the integral of (6x^2+2x-1) ?

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

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integral of (6x^2+2x-1)

\int (6 x^{2} + 2 x - 1) d x

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

Solution steps

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\frac{( 2 s + 12 )}{s ^{2} + 2 s + 5}

\frac{\partial}{\partial x} ((x - 6) (y + 4))

\frac{d}{d x} (\frac{8 x}{7 - tan ( x )})

\frac{d y}{d t} = - 2 t y + 4 e^{- t^{2}}

\frac{d}{d x} (x^{\frac{2}{3}} (\frac{5}{2} - x))

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