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A perfect square trinomial is a special type of quadratic expression that can be written as a square of a binomial. These expressions take the form \((a + b)^2\) or \((a - b)^2\). You'll often find that perfect square trinomials appear in various mathematical problems, simplifying complex calculations.
In the given exercise, the quadratic expression \(x^2 - 6x + 9\) is an example of a perfect square trinomial because it can be rewritten as \((x-3)^2\). Here’s a breakdown:

• The term \(x^2\) is equivalent to \(x \times x\).
• The term \(-6x\) comes from \(-3 \times x\) added twice (since \(2\times a \times b\)).
• The constant \(9\) is \(-3\times -3\).

Rewriting quadratics as perfect squares is useful in calculus, especially when integrating, since it helps identify a simpler integral form that is easier to handle with established formulas.

The substitution method is a powerful tool for solving integrals, particularly when the integral resembles a standard form. In this case, we'll simplify by utilizing substitution to facilitate easier integration.
In the exercise, once the quadratic \(x^2 - 6x + 9\) was rewritten as \((x-3)^2\), it was apparent that the integral can be addressed using substitution. Although a sophisticated substitution wasn't necessary here because of the direct resemblance to a known form, recognizing this enabled the application of a standard integral result straightaway.
Here's how the substitution method makes this simpler:

• Identify a section of the integral's formula that you can substitute with a single variable.
• Substitution can often transform a difficult integral into an easily solvable one.
• In this case, notice that \((x-3)^2\) quickly leads to an integral that follows the rule for \( \int \frac{1}{x-a} \, dx \).

This example shows how recognizing a structure similar to a standard form can directly lead to substitution, thereby streamlining the solution process.

Evaluating definite integrals involves determining the net area under a curve over a given interval on the x-axis. Integrals are vital in calculus for solving a wide array of problems, from simple area computations to complex physics applications.
In the given problem, after identifying that the integral resembles the form \( \int \frac{dx}{(x-a)^2} \), the function is integrated using the antiderivative, which is \(-\frac{1}{x-a}+C\).
For definite integrals, bounds are applied after integrating:

• Calculate the antiderivative function.
• Evaluate it at the upper and lower bounds.
• Subtract the lower bound result from the upper bound result to find the solution.

For this integral \( \int_{1}^{2} \frac{dx}{(x-3)^2} \), applying the bounds gives:

• Evaluate \(-\frac{1}{x-3}\) at \(x=2\) and \(x=1\).
• Subtract results: \(-\frac{1}{-1} - \left( -\frac{1}{-2} \right) = 1 - \frac{1}{2} = \frac{1}{2}\).

This demonstrates how knowing standard integral forms and accurately applying limits can make evaluating definite integrals straightforward.