What is the integral of 3x^2+8x-9 ?

The integral of 3x^2+8x-9 is x^3+4x^2-9x+C

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integral of 3x^2+8x-9

\int 3 x^{2} + 8 x - 9 d x

x^{3} + 4 x^{2} - 9 x + C

Solution steps

\int 3 x^{2} + 8 x - 9 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 8 x d x - \int 9 d x

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

\int 8 x d x = 4 x^{2}

\int 9 d x = 9 x

= x^{3} + 4 x^{2} - 9 x

Add a constant to the solution

= x^{3} + 4 x^{2} - 9 x + C

\frac{\partial}{\partial x} (ln (t + 2 c x))

f (x) = a x^{b} + d e^{- c x}

\int (4^{x} - sec^{2} (x)) d x

\int \frac{d}{1} (e^{- x} y) d x

\frac{d y}{d x} = \frac{x + y ^{2}}{2 y}, y (0) = 1