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 x} (x y)

(4 x + y e^{\frac{y}{x}}) d x - x e^{\frac{y}{x}} d y = 0

(e^{- x}) (x^{2} - x)

\int \frac{( x ^{2} )}{9 + x ^{2}} d x