The problem is asking to compute the improper integral

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

or prove its divergence/convergence.

Step-by-Step Solution:

1. Identify the singularity:
   The integrand $\frac{1}{(x+3)^2}$ becomes undefined at $x = -3$, making it an improper integral.

   Thus, we must split the integral at the singular point $x = -3$ and compute it as a limit. The integral becomes: $\int_{-3}^2 \frac{dx}{(x+3)^2} = \lim_{\epsilon \to 0^+} \left[ \int_{-3-\epsilon}^{-3} \frac{dx}{(x+3)^2} + \int_{-3}^{2} \frac{dx}{(x+3)^2} \right].$

2. Compute the antiderivative:
   The general form of the integral is: $\int \frac{dx}{(x+3)^2} = -\frac{1}{x+3} + C.$

3. Evaluate the limits of integration:

   • First Part ($\int_{-3-\epsilon}^{-3}$): $\int_{-3-\epsilon}^{-3} \frac{dx}{(x+3)^2} = \lim_{\epsilon \to 0^+} \left[ -\frac{1}{x+3} \bigg|_{-3-\epsilon}^{-3} \right].$ Compute the bounds: $-\frac{1}{x+3} \bigg|_{-3-\epsilon}^{-3} = -\frac{1}{-3 + 3} + \frac{1}{(-3-\epsilon)+3}.$ The term $-\frac{1}{-3+3}$ diverges to $-\infty$ because $x = -3$ is a vertical asymptote. Hence, the first integral diverges.

4. Conclusion:
   Since the first part of the integral diverges, the entire integral is divergent.

Would you like a further breakdown of the steps?
Here are some related questions for practice:

1. How can you determine if an improper integral converges or diverges?
2. What is the difference between improper integrals of the first and second kind?
3. How does the behavior of $\frac{1}{(x+3)^2}$ at $x = -3$ influence convergence?
4. Can improper integrals have finite values if their integrand has a singularity?
5. How would the integral change if the bounds were $-2$ to $2$?

Tip: When working with improper integrals, always split the integral at singular points and analyze convergence carefully!