Problem 85 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int \frac{3-x}{x^{2}-9} d x $$

Step-by-Step Solution

Answer

\( \int \frac{3-x}{x^2-9} \, dx = \frac{1}{2} \ln \left( \frac{|x-3|}{|x+3|} \right) + C \).

Step 1: Recognize the Integral Type

The given integral is \( \int \frac{3-x}{x^{2}-9} \ dx \). This looks like a rational function that may require partial fraction decomposition to simplify the integration process.

Step 2: Factor the Denominator

The denominator \( x^2 - 9 \) can be factored as \( (x-3)(x+3) \). This factorization will help us set up the partial fraction decomposition.

Step 3: Set Up Partial Fraction Decomposition

Express \( \frac{3-x}{x^2-9} \) as a sum of fractions: \( \frac{3-x}{(x-3)(x+3)} = \frac{A}{x-3} + \frac{B}{x+3} \). Here, \( A \) and \( B \) are constants that need to be determined.

Step 4: Solve for the Coefficients A and B

Multiply both sides by \( (x-3)(x+3) \) to clear the denominators: \[ 3-x = A(x+3) + B(x-3) \].Expand and equate coefficients:\[ 3-x = Ax + 3A + Bx - 3B \].\(Ax + Bx = 0\) implies \(A + B = 0\), and \(3A - 3B = 3\). Solving these equations, we substitute \(A = -B\) in the second equation:\[ 3(-B) - 3B = 3 \].\[-6B = 3 \]\[B = -\frac{1}{2} \]\[A = \frac{1}{2} \].

Step 5: Rewrite the Integral Using A and B

Substitute \( A = \frac{1}{2} \) and \( B = -\frac{1}{2} \) back into the partial fractions:\[ \frac{3-x}{x^2-9} = \frac{1/2}{x-3} - \frac{1/2}{x+3} \].Now express the integral as:\[ \int \left( \frac{1/2}{x-3} - \frac{1/2}{x+3} \right) dx \].

Step 6: Integrate Each Term

Integrate each term separately:\[ \int \frac{1/2}{x-3} \, dx - \int \frac{1/2}{x+3} \, dx \].The antiderivative of \( \frac{1/2}{x-a} \) is \( \frac{1}{2}\ln|x - a| \).Thus, the integral becomes:\[ \frac{1}{2} \ln |x-3| - \frac{1}{2} \ln |x+3| + C \].

Step 7: Combine the Logarithms Using Log Properties

Use the property \( \ln a - \ln b = \ln \left( \frac{a}{b} \right) \) to combine the logarithms:\[ \frac{1}{2} \ln \left( \frac{|x-3|}{|x+3|} \right) + C \].

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a powerful technique to simplify the integration of rational functions. A rational function is a fraction where both the numerator and the denominator are polynomials. To make these functions easier to integrate, we can express them as a sum of simpler fractions, called partial fractions.
Before starting with decomposition, ensure the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator. If not, perform polynomial division first.

To decompose the rational function, factor the denominator into its irreducible components. For instance, for the denominator \(x^2 - 9\) in the given integral, we factor it as \((x-3)(x+3)\). Then, express the original rational function as a sum of fractions with these factors as their denominators:

• \(\frac{A}{x-3}+\frac{B}{x+3}\), where \(A\) and \(B\) are constants.

Next, solve for \(A\) and \(B\) by clearing the denominators and equating coefficients. By expanding and equating, we can compare the polynomial's coefficients to find the values of these constants.

Rational Function Integration

Integrating rational functions often requires breaking down complex expressions into simpler terms. Once we achieve partial fraction decomposition, each term becomes easier to manage. This transformation leverages the linearity of integrals, which states that the integral of a sum is the sum of the integrals.

Consider the expression \(\int \frac{A}{x-a}\,dx + \int \frac{B}{x-b}\,dx\). Each term is now a simpler integral of the form \(\int \frac{k}{x-c}\,dx\), with \(k\) being a constant.

• Each of these types of integrals follows a standard form: the antiderivative of \(\frac{k}{x-c}\) is \(k \ln |x-c|\).

This reformulation is especially powerful because it reduces the original problem to a set of basic integrals that can be easily solved. This approach allows students to bypass cumbersome algebraic manipulations and directly apply integral formulae.

Logarithmic Properties in Integration

Logarithmic properties can simplify results in integral calculus, particularly when combining terms. In our specific exercise, after integrating using partial fractions, we encountered the solution:

\[ \frac{1}{2} \ln |x-3| - \frac{1}{2} \ln |x+3| + C \]

Here, we can apply the logarithmic identity:

• \( \ln a - \ln b = \ln \left( \frac{a}{b} \right) \)

Using this property, we can combine the two logarithmic terms into:

\[ \frac{1}{2} \ln \left( \frac{|x-3|}{|x+3|} \right) + C \]This step not only condenses the expression but also presents the result in a more simplified and recognizable form. Emphasizing these properties highlights the powerful role logarithms play in simplifying and managing expressions that arise in the integration of rational functions. This is crucial for students to understand how to handle results with logarithmic expressions efficiently.