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

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

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integral of (3x^2+4x-6)

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

x^{3} + 2 x^{2} - 6 x + C

Solution steps

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

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

\int 4 x d x = 2 x^{2}

\int 6 d x = 6 x

= x^{3} + 2 x^{2} - 6 x

Add a constant to the solution

= x^{3} + 2 x^{2} - 6 x + C

\frac{d}{d x} (- \frac{1}{3} (x^{- 3} - x^{6}))

33 + 4 t

\frac{x ^{2}}{( 4 + x )}

\int \frac{1}{1 + cos ( θ )} d θ

\frac{\partial}{\partial x} (3 x^{2} cos^{2} (x - y))

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