The given integral is:

$\int_{\frac{2\sqrt{3}}{3}}^{\infty} \frac{dx}{x^2 + 4}.$

Solution:

This is a definite integral of the form:

$\int \frac{1}{x^2 + a^2} dx,$

where $a = 2$. The standard formula for such integrals is:

$\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C.$

We will apply this formula, evaluate at the bounds, and compute the result.

Step 1: General Solution

Substituting $a = 2$, the indefinite integral becomes:

$\int \frac{1}{x^2 + 4} dx = \frac{1}{2} \arctan\left(\frac{x}{2}\right) + C.$

Step 2: Evaluate the Definite Integral

We now evaluate the definite integral from $x = \frac{2\sqrt{3}}{3}$ to $x = \infty$:

$\int_{\frac{2\sqrt{3}}{3}}^{\infty} \frac{dx}{x^2 + 4} = \left[\frac{1}{2} \arctan\left(\frac{x}{2}\right)\right]_{\frac{2\sqrt{3}}{3}}^{\infty}.$

Step 3: Evaluate at the Upper Limit ($x = \infty$)

The arctangent function satisfies:

$\arctan(x) \to \frac{\pi}{2} \quad \text{as } x \to \infty.$

Thus:

$\frac{1}{2} \arctan\left(\frac{\infty}{2}\right) = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}.$

Step 4: Evaluate at the Lower Limit ($x = \frac{2\sqrt{3}}{3}$)

Substitute $x = \frac{2\sqrt{3}}{3}$:

$\frac{1}{2} \arctan\left(\frac{\frac{2\sqrt{3}}{3}}{2}\right) = \frac{1}{2} \arctan\left(\frac{\sqrt{3}}{3}\right).$

The value of $\arctan\left(\frac{\sqrt{3}}{3}\right)$ is $\frac{\pi}{6}$. Therefore:

$\frac{1}{2} \arctan\left(\frac{\frac{2\sqrt{3}}{3}}{2}\right) = \frac{1}{2} \cdot \frac{\pi}{6} = \frac{\pi}{12}.$

Step 5: Final Answer

Now subtract the values at the bounds:

$\int_{\frac{2\sqrt{3}}{3}}^{\infty} \frac{dx}{x^2 + 4} = \frac{\pi}{4} - \frac{\pi}{12}.$

Simplify:

$\frac{\pi}{4} - \frac{\pi}{12} = \frac{3\pi}{12} - \frac{\pi}{12} = \frac{2\pi}{12} = \frac{\pi}{6}.$

Final Result:

$\boxed{\frac{\pi}{6}}$

Would you like a more detailed explanation of any step or related concepts?

Follow-up Questions:

1. How do you identify whether an improper integral converges or diverges?
2. What is the geometric meaning of the arctangent function?
3. How does this problem relate to the concept of limits in calculus?
4. Could you compute this integral using a substitution method? If so, how?
5. What happens if we change the bounds or the denominator in the integral?

Tip:

Always check if an integral converges before attempting to evaluate improper integrals.