It seems like you're asking to integrate the expression $\frac{x^2 - 4}{\sqrt{x}} \, dx$. I'll break it down and solve it step by step.

First, simplify the expression:

$\frac{x^2 - 4}{\sqrt{x}} = \frac{x^2}{\sqrt{x}} - \frac{4}{\sqrt{x}}$

Now, rewrite each term using exponents:

• $x^2 = x^{2}$
• $\sqrt{x} = x^{1/2}$

This leads to:

$\frac{x^2}{\sqrt{x}} = x^{2 - 1/2} = x^{3/2}$ $\frac{4}{\sqrt{x}} = 4x^{-1/2}$

Thus, the integral becomes:

$\int \left( x^{3/2} - 4x^{-1/2} \right) \, dx$

Now integrate each term:

1. $\int x^{3/2} \, dx = \frac{x^{5/2}}{5/2} = \frac{2x^{5/2}}{5}$
2. $\int 4x^{-1/2} \, dx = 4 \cdot \frac{x^{1/2}}{1/2} = 8x^{1/2}$

So, the final result is:

$\frac{2x^{5/2}}{5} + 8x^{1/2} + C$

Where $C$ is the constant of integration.

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Would you like a more detailed explanation of any part of the solution? Or do you have further questions on this?

Here are some related questions for deeper exploration:

1. What happens if we substitute a specific value for $x$ in the integral?
2. How would the solution change if the denominator was something other than $\sqrt{x}$?
3. How do integrals with fractional exponents behave geometrically?
4. What is the integral of $\frac{1}{x^{3/2}}$?
5. How does the constant of integration $C$ affect the result?

Tip: Always simplify an expression as much as possible before trying to integrate. It can make the process much smoother!