Problem 8 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(5-14,\) evaluate the integral. $$\int \frac{x^{2}-6}{x^{2}-9} d x$$

Step-by-Step Solution

Answer

The solution of the integral \(\int \frac{x^{2}-6}{x^{2}-9} dx\) is \(x + \frac{1}{2} \ln | \frac{x-3}{x+3} | + C\).

Step 1: Polynomial Division

Using polynomial division (or by observation), one can write \(\frac{x^{2}-6}{x^{2}-9}\) as \(1 + \frac{3}{x^{2}-9}\). Therefore, the integral becomes \(\int [1 + \frac{3}{x^{2}-9}] dx\).

Step 2: Splitting the integral

Split the integral into 2 parts: \(\int [1 + \frac{3}{x^{2}-9}] dx = \int dx + 3 \int \frac{dx}{x^{2}-9}\).

Step 3: Evaluate the integrals

The integral of 1 with respect to x is x. The 2nd integral can be solved by trigonometric substitution or recognizing it as a standard form. The standard form is \(\int \frac{dx}{x^{2}-a^{2}} = \frac{1}{2a} \ln | \frac{x-a}{x+a} | + C\). Therefore, \(\int \frac{dx}{x^{2}-9} = \frac{1}{6} \ln | \frac{x-3}{x+3} | + C\). Sum up the two integrals to get the final solution.

Key Concepts

Polynomial Division

When dealing with integrals involving rational expressions, polynomial division can often simplify the process. Polynomial division helps to break down complex fractions into simpler parts. In this exercise, we have the fraction \(\frac{x^{2}-6}{x^{2}-9}\). By applying polynomial division, or observing the expression, we can rewrite it as \(1 + \frac{3}{x^{2}-9}\). Once in this form, the integral becomes much easier to manage. This simplification turns the problem into shorter and more manageable integrals for each part of the expression.

• Helps simplify rational expressions
• Breaks complex fractions into sums of simpler fractions
• Makes the integration process more straightforward

Understanding polynomial division is crucial in calculus, as it allows you to handle and reduce fractions before integration.

Integral Calculus

Integral calculus focuses on deriving the total size or value, such as area under curves. It consists of determining antiderivatives. In this exercise, we split the integral given after polynomial division. So, \(\int [1 + \frac{3}{x^{2}-9}] dx\) becomes two separate integrals: \(\int dx\) and \(3 \int \frac{dx}{x^{2}-9}\).

• Divides complex integrals into simpler parts
• Allows for easier computation of each part

Working with separate integrals streamlines the computation process. The integral of 1 with respect to \(x\) is simply \(x\), a straightforward calculation. The more complex integral \(\int \frac{dx}{x^{2}-9}\) requires additional techniques, such as trigonometric substitution or applying a standard integral form. These strategies will be explored in further sections.

Trigonometric Substitution

Trigonometric substitution is a technique used to evaluate integrals involving square roots or quadratic expressions. Although not directly used in our exercise due to the advantage of a standard form integration, it's a powerful method when needed. Suppose we deal with \(\int \frac{dx}{x^{2} - a^{2}}\), substitutions like \(x = a \sec(\theta)\) or \(x = a \tan(\theta)\) replace the variable \(x\) with a trigonometric function.

• Useful for integrals with quadratic expressions
• Simplifies the integral by changing the variable
• Often involves new variable and angle calculations

While trigonometric substitution wasn't necessary here, it remains a valuable tool. Recognizing when to apply it can simplify problems involving complex square roots or quadratic expressions.

Standard Form Integral

In our exercise, recognizing a standard form integral formula significantly simplifies solving \(\int \frac{dx}{x^{2}-9}\). The integral \(\int \frac{dx}{x^{2}-a^{2}}\) is a standard calculus form, allowing a direct application of a known result:\[\int \frac{dx}{x^{2}-a^{2}} = \frac{1}{2a} \ln \left| \frac{x-a}{x+a} \right| + C\]In our case, substituting \(a = 3\), turns the integral into \(\frac{1}{6} \ln \left| \frac{x-3}{x+3} \right| + C\). This approach saves time and reduces errors by relying on established methods:

• Applies known results from calculus
• Avoids unnecessary substitutions or transformations
• Efficiently handles rational expressions of certain types

Mastering standard form integrals allows for quick and accurate integral evaluations, crucial in calculus problems.