The problem asks us to evaluate the improper integral:

$\int_{-3}^{2} \frac{dx}{(x + 3)^2}$

or prove its divergence (or convergence). Let us solve it step by step.

Step 1: Analyze the function and detect possible singularities

The function to integrate is:

$f(x) = \frac{1}{(x + 3)^2}.$

• The denominator $(x + 3)^2$ becomes zero at $x = -3$, which means $x = -3$ is a singular point.
• Thus, the integral is improper due to the singularity at $x = -3$. We will need to analyze it as a limit to check for convergence or divergence.

Step 2: Rewrite the integral as a limit

The improper integral is rewritten as:

$\int_{-3}^{2} \frac{dx}{(x + 3)^2} = \lim_{\varepsilon \to 0^+} \left( \int_{-3 + \varepsilon}^{2} \frac{dx}{(x + 3)^2} \right).$

Step 3: Evaluate the antiderivative of $\frac{1}{(x + 3)^2}$

We calculate the indefinite integral of $\frac{1}{(x + 3)^2}$:

$\int \frac{1}{(x + 3)^2} dx = -\frac{1}{x + 3} + C.$

Step 4: Substitute the limits

Using the antiderivative, the integral becomes:

$\int_{-3 + \varepsilon}^{2} \frac{dx}{(x + 3)^2} = \left[ -\frac{1}{x + 3} \right]_{-3 + \varepsilon}^{2}.$

Substitute the bounds: $\left[ -\frac{1}{x + 3} \right]_{-3 + \varepsilon}^{2} = -\frac{1}{2 + 3} - \left(-\frac{1}{(-3 + \varepsilon) + 3}\right).$

Simplify: $= -\frac{1}{5} - \left(-\frac{1}{\varepsilon}\right).$ $= -\frac{1}{5} + \frac{1}{\varepsilon}.$

Step 5: Analyze the limit as $\varepsilon \to 0^+$

As $\varepsilon \to 0^+$, the term $\frac{1}{\varepsilon}$ tends to infinity. Thus, the integral diverges because the singularity at $x = -3$ causes the value to blow up.

Conclusion

The improper integral $\int_{-3}^{2} \frac{dx}{(x + 3)^2}$ diverges.

Would you like further clarification or details on any step? Let me know!
Here are 5 related questions to expand your understanding:

1. What are the different types of improper integrals, and how do we analyze their convergence?
2. How would the solution change if the power in the denominator were $(x + 3)^p$, where $p > 2$?
3. What are common techniques to evaluate improper integrals with infinite limits?
4. Can you explain why $\frac{1}{\varepsilon}$ causes divergence as $\varepsilon \to 0^+$?
5. How can improper integrals be applied to real-world problems, such as physics or probability?

Tip: When analyzing improper integrals, always check for singularities or infinite bounds before attempting to evaluate them.