What is the integral of (e^{3x+9}-3x^2) ?

The integral of (e^{3x+9}-3x^2) is 1/3 e^{3x+9}-x^3+C

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integral of (e^{3x+9}-3x^2)

\int (e^{3 x + 9} - 3 x^{2}) d x

\frac{1}{3} e^{3 x + 9} - x^{3} + C

Solution steps

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

\int e^{3 x + 9} d x = \frac{1}{3} e^{3 x + 9}

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

= \frac{1}{3} e^{3 x + 9} - x^{3}

Add a constant to the solution

= \frac{1}{3} e^{3 x + 9} - x^{3} + C

y = 8 arcsin (\frac{x}{4}) - \frac{x 16 - x ^{2}}{2}

\frac{\partial}{\partial x} (e^{y} cos (x))

\int_{- 1}^{1} 9 x^{2} + 6 x - 3 d x

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

\frac{d}{d x} (\frac{1}{1 + x + x ^{2}})

What is the integral of (e^{3x+9}-3x^2) ?
The integral of (e^{3x+9}-3x^2) is 1/3 e^{3x+9}-x^3+C