To integrate the function $f(x) = 6x^4 - 9x^2 - \sqrt{x}$, we can handle each term separately. Let's compute the integral step by step.

1. Integral of $6x^4$:

$\int 6x^4 \, dx = 6 \cdot \frac{x^5}{5} = \frac{6x^5}{5}$

2. Integral of $-9x^2$:

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

3. Integral of $-\sqrt{x}$:

We rewrite $\sqrt{x}$ as $x^{1/2}$, then integrate:

$\int -x^{1/2} \, dx = - \frac{x^{3/2}}{3/2} = -\frac{2}{3}x^{3/2}$

Putting it all together:

The indefinite integral of $f(x) = 6x^4 - 9x^2 - \sqrt{x}$ is:

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

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 of these steps?

Related Questions:

1. How do you find the definite integral of this function over a specific interval?
2. How do you integrate functions involving fractional exponents?
3. What are common mistakes when integrating polynomial and radical functions?
4. How can you verify if an integral has been computed correctly?
5. Can you apply substitution or other techniques to simplify this integral?

Tip:

Always simplify radical expressions before integrating to avoid errors, especially when dealing with fractional powers like $\sqrt{x} = x^{1/2}$.