The problem asks to compute the following integral:

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

To solve this, we can approach the two terms separately.

Step 1: Solve the first integral

We start by calculating the integral of the first term:

$I_1 = \int_0^\infty \frac{1}{\sqrt{x^2 + 4}} dx$

This is a standard integral and can be solved by using the substitution $x = 2 \tan(\theta)$, which simplifies the expression. After performing the substitution, we know that the result is:

$I_1 = \ln\left(\frac{\sqrt{x^2 + 4} + x}{2}\right)\bigg|_0^\infty = \ln(2)$

Step 2: Solve the second integral

Now, we calculate the second integral:

$I_2 = \int_0^\infty \frac{1}{x + 2} dx$

This is another standard integral. The result of this integral is:

$I_2 = \ln(x + 2)\bigg|_0^\infty = \ln(\infty) - \ln(2) = \infty$

Step 3: Combine the two results

We now subtract the two results:

$I = I_1 - I_2 = \ln(2) - \infty$

Since subtracting infinity leads to an undefined result, this integral diverges.

Thus, the integral does not converge to a finite value.

Would you like more details or have any questions?

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Relative Questions:

1. How is the substitution $x = 2 \tan(\theta)$ applied in integrals?
2. Can integrals of rational functions always be solved with logarithmic terms?
3. What are conditions for an improper integral to converge?
4. How can the divergence of an integral be interpreted?
5. What methods exist for handling divergent integrals?

Tip: When dealing with improper integrals, check the behavior at the limits of integration to determine convergence.