To solve the integral:

$\int \frac{dx}{x - \sqrt{x}},$

we will first try to simplify the integrand. The expression involves both $x$ and $\sqrt{x}$, which suggests a substitution method might help.

Step 1: Make a substitution

Let:

$u = \sqrt{x} \quad \text{so that} \quad x = u^2 \quad \text{and} \quad dx = 2u \, du.$

Now, substitute $x = u^2$ and $dx = 2u \, du$ into the integral:

$\int \frac{dx}{x - \sqrt{x}} = \int \frac{2u \, du}{u^2 - u}.$

Step 2: Simplify the integrand

Factor the denominator:

$u^2 - u = u(u - 1),$

so the integral becomes:

$\int \frac{2u \, du}{u(u - 1)} = \int \frac{2 \, du}{u - 1}.$

Step 3: Integrate

Now, the integral simplifies to a basic form:

$\int \frac{2 \, du}{u - 1} = 2 \ln|u - 1| + C.$

Step 4: Back-substitute $u = \sqrt{x}$

Finally, substitute $u = \sqrt{x}$ back into the result:

$2 \ln|\sqrt{x} - 1| + C.$

Thus, the solution to the integral is:

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

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Would you like further details on any of the steps?

Here are some related questions you might find interesting:

1. How can we solve integrals involving rational functions?
2. What are some other common substitution methods in integration?
3. How do you integrate a rational function with square roots in both the numerator and denominator?
4. How does logarithmic differentiation help in solving integrals?
5. How can you recognize when to use substitution in an integral?

Tip: When dealing with square roots or expressions that suggest a relationship between $x$ and $\sqrt{x}$, substitution is often a good strategy to simplify the integral.