Problem 86 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int \frac{4-x}{x^{2}-16} d x $$

Step-by-Step Solution

Answer

\[ \int \frac{4-x}{x^2-16} \, dx = \frac{1}{2}\ln\left(\frac{|x-4|}{|x+4|}\right) + C \]

Step 1: Identify the form

The integrand \( \frac{4-x}{x^2-16} \) suggests using partial fraction decomposition. The denominator \( x^2 - 16 \) can be factored as \( (x-4)(x+4) \).

Step 2: Set up the partial fraction decomposition

Express \( \frac{4-x}{(x-4)(x+4)} \) as the sum: \( \frac{A}{x-4} + \frac{B}{x+4} \). Determine the constants \( A \) and \( B \).

Step 3: Solve for constants A and B

Multiply both sides by the denominator \((x-4)(x+4)\) to clear the fractions: \( 4-x = A(x+4) + B(x-4) \). Expand and collect like terms: \( 4-x = (A+B)x + (4A - 4B) \). Set coefficients equal: \( A+B = 0 \) and \( 4A - 4B = 4 \).

Step 4: Calculate values of A and B

From the equations \( A+B = 0 \) and \( 4A - 4B = 4 \), solve for \( A \) and \( B \):1. \( A = -B \) (from \( A + B = 0 \))2. Substitute \( -B \) for \( A \) in the second equation: \( 4(-B) - 4B = 4 \) Simplify: \(-4B - 4B = 4 \) Thus, \( -8B = 4 \), so \( B = -\frac{1}{2} \).Since \( A = -B \), \( A = \frac{1}{2} \).

Step 5: Write the decomposed form of the integrand

Substitute back the terms found into the partial fractions: \( \frac{4-x}{x^2-16} = \frac{1/2}{x-4} - \frac{1/2}{x+4} \).

Step 6: Integrate each term separately

Integrate the equation term by term: \( \int \frac{1/2}{x-4}dx - \int \frac{1/2}{x+4}dx \). The integral of \( \frac{1}{x-a} \) is \( \ln|x-a| \): Thus, \( \frac{1}{2}\ln|x-4| - \frac{1}{2}\ln|x+4| + C \).

Step 7: Simplify the expression

Combine the logarithms using the properties of logarithms:\[ \frac{1}{2}\ln\left(\frac{|x-4|}{|x+4|}\right) + C \]This expression combines the two logarithmic terms into one.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a powerful technique used for simplifying rational expressions. This technique can transform complex fractions into simpler ones, which are easier to integrate or differentiate. In our exercise, we start with the integrand \( \frac{4-x}{x^2-16} \). First, we recognize that the denominator can be factored into \((x-4)(x+4)\). By expressing the fraction in terms of these factors, we can split the original fraction into simpler parts.

The goal is to write \( \frac{4-x}{(x-4)(x+4)} \) as a sum of the form \( \frac{A}{x-4} + \frac{B}{x+4} \), where \( A \) and \( B \) are constants to be determined. After setting up this equation, we clear the fractions by multiplying both sides by \((x-4)(x+4)\), leading to the equation: \( 4-x = A(x+4) + B(x-4) \).

This method is beneficial because it breaks down complex expressions into forms that can be more directly addressed through standard calculus techniques, such as logarithmic integration, as we see in the steps that follow.

Integration Techniques

Integration techniques are methods used to evaluate integrals, especially when dealing with complex expressions. One common technique is using partial fraction decomposition, as seen in our exercise. This allows us to transform the integrand into simpler terms, which can often be integrated directly.

Consider the simplified result from partial fraction decomposition: \( \frac{1/2}{x-4} - \frac{1/2}{x+4} \). Integrating each term separately is more straightforward. We use the fact that the integral of \( \frac{1}{x-a} \) is \( \ln|x-a| \). For each term, we apply this rule:

• \( \int \frac{1/2}{x-4}\,dx = \frac{1}{2}\ln|x-4| \)
• \( \int \frac{1/2}{x+4}\,dx = \frac{1/2}\ln|x+4| \)

By addressing each simpler component independently, the overall task of integration becomes manageable. These techniques are crucial in working through complex integrals efficiently and are widely applicable across many areas of mathematics and engineering.

Logarithmic Integration

Logarithmic integration is a specific integration technique that is especially useful when dealing with functions involving fractions like \( \frac{1}{x-a} \). This method uses the fact that the integral of such a function is the natural logarithm of the absolute value of the denominator: \( \ln|x-a| \).

In our exercise, after partial fraction decomposition, we arrived at two terms: \( \frac{1/2}{x-4} \) and \( -\frac{1/2}{x+4} \). For each of these, we use logarithmic integration:

• The integral of \( \frac{1/2}{x-4} \) is \( \frac{1}{2}\ln|x-4| \).
• The integral of \( -\frac{1/2}{x+4} \) is \( -\frac{1}{2}\ln|x+4| \).

After computing these integrals, they can be combined using properties of logarithms, specifically the quotient rule. This results in
\[ \frac{1}{2}\ln\left(\frac{|x-4|}{|x+4|}\right) + C \]
which consolidates the two separate logarithmic expressions into one simplified form. This is an elegant outcome of the application of logarithmic integration principles.