Problem 83 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The integral is \( \frac{1}{6} \ln|x-3| + \frac{5}{6} \ln|x+3| + C \).

Step 1: Recognize the Form for Partial Fraction Decomposition

The given integral involves a rational function, so we identify it as a candidate for partial fraction decomposition. The denominator \( x^2 - 9 \) can be factored as \( (x - 3)(x + 3) \).

Step 2: Set Up the Partial Fraction Decomposition

Express \( \frac{x+3}{x^2-9} \) as the sum of two simpler fractions: \( \frac{A}{x-3} + \frac{B}{x+3} \).

Step 3: Solve for Coefficients A and B

To find \( A \) and \( B \), multiply through by the common denominator, obtaining \( x + 3 = A(x + 3) + B(x - 3) \). Simplify and equate coefficients to solve for \( A \) and \( B \).

Step 4: Substitute Back into the Integral

Using \( A = \frac{1}{6} \) and \( B = \frac{5}{6} \) from prior calculations, rewrite the integral: \[\int \left( \frac{1}{6(x-3)} + \frac{5}{6(x+3)} \right) dx\]

Step 5: Integrate Each Term Separately

Integrate each term separately as:\( \int \frac{1}{6(x-3)} dx = \frac{1}{6} \ln|x-3| + C_1 \) \( \int \frac{5}{6(x+3)} dx = \frac{5}{6} \ln|x+3| + C_2 \)

Step 6: Combine the Integrals

Combine the results from Step 5 into a single expression:\[\frac{1}{6} \ln|x-3| + \frac{5}{6} \ln|x+3| + C\]where \( C = C_1 + C_2 \) is an arbitrary constant.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in calculus to break down complex rational functions into simpler fractions, making them easier to integrate. The process begins by expressing the original rational function as a sum of simpler fractions called partial fractions.
This often involves factoring the denominator of the rational function into linear terms.

• Each factor corresponds to a term in the partial fraction decomposition.
• The expression \[\frac{x+3}{x^2-9}\]can be rewritten as\[\frac{A}{x-3} + \frac{B}{x+3}.\]This is done by identifying that \( x^2 - 9 \) can be factored into \( (x-3)(x+3) \).

Next, you solve for the coefficients \( A \) and \( B \) by clearing the denominators and equating coefficients. Ultimately, these coefficients are determined by substituting convenient values for the variable \( x \) or by comparing coefficients. This allows the integrals of these simpler fractions to be more manageable, leading to easier integration methods such as using logarithmic functions.

Indefinite Integrals

Indefinite integrals are essentially the reverse process of differentiation. They are used to find a function, often referred to as the antiderivative, from which a given function could have been derived through differentiation. In its simplest form, an indefinite integral is represented as:

• \[\int f(x) \, dx = F(x) + C\]
• where \( F(x) \) is the antiderivative of \( f(x) \),
• and \( C \) is the constant of integration, representing an infinite number of potential antiderivatives.

For rational functions decomposed using partial fraction decomposition, indefinite integrals can often be solved term by term. As seen in the problem:

• The integral \[\int \frac{x+3}{x^2-9} \, dx \] was split into two simpler integrals through partial fractions \[\int \frac{1}{6(x-3)} \, dx + \int \frac{5}{6(x+3)} \, dx.\]
• Each term can be separately integrated using logarithmic properties since they present functions of the form \( \frac{1}{x-a} \).

Rational Functions

Rational functions are a crucial part of calculus and represent ratios of two polynomials. These functions are in the form:

• \( \frac{P(x)}{Q(x)} \)
  where both \( P(x) \) and \( Q(x) \) are polynomials.
• If \( Q(x) \) is zero at any value of \( x \), the function is undefined at that point.

Integrating rational functions often requires techniques like partial fraction decomposition because direct integration isn't usually straightforward. This is particularly true when the degree of the numerator is less than the degree of the denominator, as integration can then be simplified significantly. For example, in the exercise provided, since the original function \( x+3 \)is of lower degree than \( x^2-9 \), partial fraction decomposition helps convert this into a form where each part has a direct antiderivative, mostly involving simple logarithmic functions.