The integral provided is:

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

Let me help you solve this step by step.

Step 1: Simplify the powers of $x$

Recall that:

• $x^{3/2}$ is $x^{1.5}$,
• $\sqrt{x} = x^{1/2}$.

Thus, the integral becomes:

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

Step 2: Factor the denominator

Factor $x^{1/2}$ from the denominator:

$x^{1/2}(x^{1} - 1) = x^{1/2}(x - 1).$

Thus, the integral becomes:

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

Step 3: Substitution

Let $u = \sqrt{x} = x^{1/2}$, so $du = \frac{1}{2\sqrt{x}}dx$, or $dx = 2u \, du$.

Substitute into the integral:

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

The $u$'s cancel out, leaving:

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

Step 4: Recognize a standard integral

The integral $\int \frac{du}{u^2 - 1}$ is a standard result and can be solved using partial fractions.

$\frac{1}{u^2 - 1} = \frac{1}{2} \left( \frac{1}{u - 1} - \frac{1}{u + 1} \right).$

Thus, the integral becomes:

$2 \int \frac{du}{u^2 - 1} = \int \left( \frac{1}{u - 1} - \frac{1}{u + 1} \right) du.$

Step 5: Solve the integral

This is straightforward to integrate:

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

This simplifies to:

$\ln \left| \frac{u - 1}{u + 1} \right| + C.$

Step 6: Substitute back $u = \sqrt{x}$

Finally, substitute back $u = \sqrt{x}$ to get the final answer:

$\ln \left| \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right| + C.$

Thus, the solution to the integral is:

$\ln \left| \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right| + C.$

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Would you like any further explanation or details on a specific step?

5 Related Questions:

1. How can you solve other integrals involving square roots?
2. What are some common techniques for simplifying integrals using substitution?
3. How does partial fraction decomposition help in solving integrals?
4. Can you explain how to handle improper integrals?
5. What are some alternative methods for factoring the denominator of integrals?

Tip:

When encountering powers or square roots in integrals, substitution methods like $u = \sqrt{x}$ often simplify the process dramatically!