The given integral is:

$\int_{0}^{2} \frac{dx}{(x-1)^2}$

This is an improper integral because the denominator $(x-1)^2$ becomes undefined at $x = 1$, which lies within the limits of integration. To evaluate or prove its divergence, let's analyze it step by step:

Step 1: Split the Integral Around the Problem Point $x = 1$

We split the integral into two parts around $x = 1$: $\int_{0}^{2} \frac{dx}{(x-1)^2} = \int_{0}^{1} \frac{dx}{(x-1)^2} + \int_{1}^{2} \frac{dx}{(x-1)^2}$

Step 2: Analyze Each Part

(a) First Part: $\int_{0}^{1} \frac{dx}{(x-1)^2}$

Here, as $x \to 1^{-}$, the term $(x-1)^2 \to 0^+$, so the integrand blows up. We rewrite the integral using a limit: $\int_{0}^{1} \frac{dx}{(x-1)^2} = \lim_{\epsilon \to 1^{-}} \int_{0}^{\epsilon} \frac{dx}{(x-1)^2}.$

Evaluating this: $\int \frac{dx}{(x-1)^2} = -\frac{1}{x-1} + C.$ Thus: $\int_{0}^{\epsilon} \frac{dx}{(x-1)^2} = \left[ -\frac{1}{x-1} \right]_{0}^{\epsilon} = -\frac{1}{\epsilon - 1} - (-\frac{1}{0 - 1}).$

As $\epsilon \to 1^{-}$, $-\frac{1}{\epsilon - 1} \to -\infty$. Hence, the first part diverges.

(b) Second Part: $\int_{1}^{2} \frac{dx}{(x-1)^2}$

Similarly, we use a limit to analyze this integral: $\int_{1}^{2} \frac{dx}{(x-1)^2} = \lim_{\delta \to 1^{+}} \int_{\delta}^{2} \frac{dx}{(x-1)^2}.$

Evaluating the same antiderivative: $\int_{\delta}^{2} \frac{dx}{(x-1)^2} = \left[ -\frac{1}{x-1} \right]_{\delta}^{2} = -\frac{1}{2-1} - (-\frac{1}{\delta - 1}).$

As $\delta \to 1^{+}$, $-\frac{1}{\delta - 1} \to -\infty$. Hence, the second part also diverges.

Conclusion

Since both parts of the integral diverge, the given improper integral does not converge. The integral diverges.

Would you like further clarification or details about any step? 😊

Related Questions:

1. What is an improper integral, and how is it defined?
2. How can we determine convergence of improper integrals in general?
3. Can the divergence of $\frac{1}{(x-1)^2}$ be shown using comparison with simpler functions?
4. How would this integral change if the limits avoided the singularity (e.g., $\int_{0}^{0.5} + \int_{1.5}^{2}$)?
5. How does the order of growth of $\frac{1}{(x-1)^2}$ contribute to divergence?

Tip:

Always examine the behavior of the integrand near singular points carefully before assuming convergence.