To find the integral of the function $f(x) = 6x^4 - 9x^2 - \sqrt{x}$, let's integrate each term separately.

Step-by-step integration:

1. First term: $6x^4$ $\int 6x^4 \, dx = 6 \cdot \frac{x^5}{5} = \frac{6x^5}{5}$

2. Second term: $-9x^2$ $\int -9x^2 \, dx = -9 \cdot \frac{x^3}{3} = -3x^3$

3. Third term: $-\sqrt{x}$ Recall that $\sqrt{x} = x^{1/2}$, so: $\int -\sqrt{x} \, dx = -\int x^{1/2} \, dx = -\frac{x^{3/2}}{\frac{3}{2}} = -\frac{2}{3}x^{3/2}$

Combining all the terms:

$\int (6x^4 - 9x^2 - \sqrt{x}) \, dx = \frac{6x^5}{5} - 3x^3 - \frac{2}{3}x^{3/2} + C$

Where $C$ is the constant of integration.

Final Answer:

$\boxed{\int f(x) \, dx = \frac{6x^5}{5} - 3x^3 - \frac{2}{3}x^{3/2} + C}$

Would you like more details on any specific part?

Here are 5 related questions:

1. How do you apply the power rule for integration?
2. How does integration differ when negative powers of $x$ are involved?
3. What happens when integrating a fractional exponent like $\sqrt{x}$?
4. How can you verify the result by differentiating the integrated function?
5. What methods are used when integrating functions with roots other than square roots?

Tip: When integrating functions with fractional exponents like $\sqrt{x}$, rewrite the root as a power of $x$ for easier handling.