To solve the given integral

$\int \frac{-5}{x\sqrt{x^2 - 4}} \, dx,$

we use the standard technique involving substitution for integrals containing $\sqrt{x^2 - a^2}$.

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Step 1: Identify the form

This integral suggests the substitution: $x = 2 \sec \theta, \quad dx = 2 \sec \theta \tan \theta \, d\theta, \quad \sqrt{x^2 - 4} = 2 \tan \theta.$

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Step 2: Substitute

Using the substitutions: $\frac{-5}{x\sqrt{x^2 - 4}} \, dx = \frac{-5}{2\sec\theta \cdot 2\tan\theta} \cdot 2\sec\theta \tan\theta \, d\theta,$ simplify: $= \frac{-5}{4\tan\theta} \cdot \tan\theta \, d\theta = \frac{-5}{4} \, d\theta.$

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Step 3: Integrate

The integral becomes: $\int \frac{-5}{4} \, d\theta = \frac{-5}{4} \theta + C.$

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Step 4: Back-substitute

Since $x = 2 \sec \theta$, we know: $\sec \theta = \frac{x}{2} \quad \text{and} \quad \theta = \sec^{-1}\left(\frac{x}{2}\right).$ Thus, the integral is: $\frac{-5}{4} \sec^{-1}\left(\frac{x}{2}\right) + C.$

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Final Answer:

$\int \frac{-5}{x\sqrt{x^2 - 4}} \, dx = \frac{-5}{4} \sec^{-1}\left(\frac{x}{2}\right) + C.$

Would you like me to break down any step further or explain the substitution method more?

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Related Questions:

1. How do you integrate $\int \frac{1}{x\sqrt{x^2 - a^2}} \, dx$?
2. What are some common substitutions used for $\sqrt{x^2 \pm a^2}$?
3. Can you explain the geometric interpretation of $\sec^{-1}(x)$?
4. How is the substitution $x = a \sec \theta$ derived for $\sqrt{x^2 - a^2}$?
5. How do you determine constants of integration in definite integrals?

Tip:

For integrals involving $\sqrt{x^2 - a^2}$, always check if a trigonometric substitution simplifies the expression efficiently.