The given integral is:

$\int \frac{dx}{x^{\frac{3}{2}} - \sqrt{x}}$

Step-by-Step Solution:

1. Simplify the expression in the denominator: Recall that $\sqrt{x} = x^{\frac{1}{2}}$. We can rewrite the denominator as:

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

2. Factor the denominator: Notice that the expression can be factored as:

   $x^{\frac{3}{2}} - x^{\frac{1}{2}} = x^{\frac{1}{2}} (x - 1)$

3. Rewrite the integral: Now, substitute this into the integral:

   $\int \frac{dx}{x^{\frac{1}{2}} (x - 1)}$

4. Substitution: To solve this, use a substitution. Let $u = \sqrt{x} = x^{\frac{1}{2}}$. Then, $x = u^2$ and $dx = 2u \, du$.

   Substitute these into the integral:

   $\int \frac{2u \, du}{u \cdot (u^2 - 1)}$

   Simplifying:

   $\int \frac{2 \, du}{u^2 - 1}$

5. Partial fraction decomposition: The denominator $u^2 - 1$ factors as $(u - 1)(u + 1)$. So, we can use partial fractions:

   $\frac{2}{(u - 1)(u + 1)} = \frac{A}{u - 1} + \frac{B}{u + 1}$

   Solving for $A$ and $B$, we get $A = 1$ and $B = 1$. Therefore:

   $\frac{2}{(u - 1)(u + 1)} = \frac{1}{u - 1} + \frac{1}{u + 1}$

6. Integrating: Now, we can integrate term by term:

   $\int \left( \frac{1}{u - 1} + \frac{1}{u + 1} \right) du = \ln|u - 1| + \ln|u + 1| + C$

7. Back-substitute $u = \sqrt{x}$:

   $\ln|\sqrt{x} - 1| + \ln|\sqrt{x} + 1| + C$

   Using properties of logarithms, this simplifies to:

   $\ln\left| (\sqrt{x} - 1)(\sqrt{x} + 1) \right| + C$

   Since $(\sqrt{x} - 1)(\sqrt{x} + 1) = x - 1$, the final answer is:

   $\ln|x - 1| + C$

Final Answer:

$\int \frac{dx}{x^{\frac{3}{2}} - \sqrt{x}} = \ln|x - 1| + C$

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Would you like any more details or explanations?

Here are some related questions for practice:

1. What is the integral of $\int \frac{dx}{\sqrt{x}(x+1)}$?
2. How can you solve $\int \frac{dx}{x^2 - 1}$ using partial fractions?
3. What substitution would you use for $\int \frac{dx}{\sqrt{x}(x-2)}$?
4. How does partial fraction decomposition work for quadratic denominators?
5. What is the role of logarithmic functions in integrals involving rational expressions?

Tip: Always look for patterns or common factorizations in complex rational functions—they can often simplify the integral dramatically.