To solve the integral

$\int \frac{3 \, dx}{x \sqrt{x^4 - 4}},$

we will change variables using a substitution method to simplify the square root. Let's proceed step by step:

Step 1: Substitution

Let: $u = x^2.$ Then: $du = 2x \, dx \quad \text{or} \quad \frac{du}{2} = x \, dx.$

The square root term becomes: $\sqrt{x^4 - 4} = \sqrt{u^2 - 4}.$

Step 2: Rewrite the Integral

Substituting into the integral: $\int \frac{3 \, dx}{x \sqrt{x^4 - 4}} = \int \frac{3}{x \sqrt{x^4 - 4}} \cdot dx = \int \frac{3}{\sqrt{u^2 - 4}} \cdot \frac{du}{2}.$

Simplify the constants: $\int \frac{3}{\sqrt{u^2 - 4}} \cdot \frac{du}{2} = \frac{3}{2} \int \frac{1}{\sqrt{u^2 - 4}} \, du.$

Step 3: Solve the Simplified Integral

The integral $\int \frac{1}{\sqrt{u^2 - 4}} \, du$ is a standard form: $\int \frac{1}{\sqrt{u^2 - a^2}} \, du = \ln|u + \sqrt{u^2 - a^2}| + C,$ where $a^2 = 4$, so $a = 2$.

Thus: $\int \frac{1}{\sqrt{u^2 - 4}} \, du = \ln|u + \sqrt{u^2 - 4}| + C.$

Step 4: Substitute Back $u = x^2$

Returning to the original variable: $\frac{3}{2} \int \frac{1}{\sqrt{u^2 - 4}} \, du = \frac{3}{2} \ln|u + \sqrt{u^2 - 4}| + C,$ and substituting $u = x^2$: $\frac{3}{2} \ln|x^2 + \sqrt{x^4 - 4}| + C.$

Final Answer:

$\int \frac{3 \, dx}{x \sqrt{x^4 - 4}} = \frac{3}{2} \ln|x^2 + \sqrt{x^4 - 4}| + C.$

Would you like me to go deeper into any part of this solution? Here are some related questions:

1. How is the substitution method derived?
2. Why is $\int \frac{1}{\sqrt{u^2 - a^2}} \, du$ a standard integral form?
3. Can we solve this without substitution?
4. How do constants affect the integral’s solution?
5. What does this integral represent geometrically?

Tip: Always simplify square roots before substitution for more manageable calculations!