To solve the integral:

$I = \int_0^\infty \left( \frac{1}{\sqrt{x^2 + 4}} - \frac{C}{x + 2} \right) dx$

we can approach each term separately and deal with potential convergence issues.

Step 1: First integral $\int_0^\infty \frac{1}{\sqrt{x^2 + 4}} dx$

This is a standard integral form, which can be computed using a trigonometric substitution.

• Let $x = 2 \tan(\theta)$, so that $dx = 2 \sec^2(\theta) d\theta$, and $\sqrt{x^2 + 4} = 2\sec(\theta)$.

This substitution transforms the integral into:

$\int_0^\infty \frac{1}{\sqrt{x^2 + 4}} dx = \int_0^{\frac{\pi}{2}} d\theta = \frac{\pi}{2}$

Thus:

$\int_0^\infty \frac{1}{\sqrt{x^2 + 4}} dx = \frac{\pi}{2}$

Step 2: Second integral $\int_0^\infty \frac{C}{x + 2} dx$

This integral diverges as $x \to \infty$. However, if we subtract these terms, the divergence may cancel out. We need to analyze this carefully. To simplify, consider this integral as a known logarithmic form:

$\int_0^\infty \frac{C}{x + 2} dx = C \cdot \lim_{a \to \infty} \ln\left( \frac{a + 2}{2} \right)$

This expression diverges logarithmically, and we'll need to adjust $C$ to manage this divergence.

Conclusion

This integral will converge for specific values of $C$ that cancel the divergence of the logarithmic term. If you provide more context (or a specific value for $C$), I can solve the integral more precisely.

Do you want more details on this method or have any specific questions?

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Relative Questions:

1. What is the value of $C$ that ensures convergence?
2. How does trigonometric substitution work in similar integrals?
3. What happens when both terms of an integral have diverging behavior?
4. How does the logarithmic term influence the result in divergent integrals?
5. Can we apply the method of improper integrals for large values of $x$?

Tip: For improper integrals, always check the limits and divergence behavior carefully before attempting to solve!