To find the integral of the function $f(x) = 3x^2 + 6x - 9$, we apply the power rule for integration to each term.

The general rule for integrating a power of $x$ is:

$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$

Now, let's integrate each term:

$\int (3x^2 + 6x - 9) \, dx = \int 3x^2 \, dx + \int 6x \, dx + \int -9 \, dx$

1. For the first term $3x^2$:

$\int 3x^2 \, dx = 3 \cdot \frac{x^{2+1}}{2+1} = 3 \cdot \frac{x^3}{3} = x^3$

2. For the second term $6x$:

$\int 6x \, dx = 6 \cdot \frac{x^{1+1}}{1+1} = 6 \cdot \frac{x^2}{2} = 3x^2$

3. For the third term $-9$:

$\int -9 \, dx = -9x$

Putting it all together:

$\int (3x^2 + 6x - 9) \, dx = x^3 + 3x^2 - 9x + C$

where $C$ is the constant of integration.

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Would you like further details or explanations?

Here are 5 related questions:

1. How do you find definite integrals?
2. What is the geometric interpretation of an integral?
3. How do you apply integration to physics problems?
4. How can you integrate a function with trigonometric terms?
5. What happens if the function has a fractional exponent?

Tip: Always remember to add the constant of integration $C$ when performing indefinite integrals!