Problem 84 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The integral is \( \ln|x-4| + C \).

Step 1: Identify the Decomposition Type

The integrand \( \frac{x+4}{x^{2}-16} \) is a rational function. To integrate it, we should decompose it into partial fractions, if possible. Notice that the denominator \( x^{2} - 16 \) is a difference of squares that can be factored as \( (x-4)(x+4) \).

Step 2: Decompose into Partial Fractions

Express \( \frac{x+4}{x^{2}-16} \) as a sum of partial fractions. Set:\[\frac{x+4}{x^2-16} = \frac{A}{x-4} + \frac{B}{x+4}\]Multiply both sides by the denominator \( x^2-16 \) to clear the fractions:\[ x+4 = A(x+4) + B(x-4) \]

Step 3: Solve for Constants A and B

Expand and combine terms on the right side:\[x + 4 = Ax + 4A + Bx - 4B\]Combine like terms:\[ x + 4 = (A + B)x + (4A - 4B)\]By equating coefficients, we get the system:1. \( A + B = 1 \)2. \( 4A - 4B = 4 \)

Step 4: Solve the System of Equations

Solve the system of equations:From equation 1, \( A = 1 - B \).Substitute into equation 2:\[ 4(1 - B) - 4B = 4 \]Simplify:\[ 4 - 4B - 4B = 4 \]\[ 4 - 8B = 4 \]Solving for \( B \):\[ -8B = 0 \]\[ B = 0 \]Substitute back to find \( A \):\( A + 0 = 1 \) thus, \( A = 1 \).

Step 5: Set Up Integral for Each Term

Having found \( A = 1 \) and \( B = 0 \), rewrite the integral:\[\int \frac{1}{x-4} \, dx + \int \frac{0}{x+4} \, dx\]Simplifying, the integral becomes:\[\int \frac{1}{x-4} \, dx\]

Step 6: Integrate Each Term Separately

The integral \( \int \frac{1}{x-4} \, dx \) can be solved using the natural logarithm:\[\int \frac{1}{x-4} \, dx = \ln|x-4| + C\]

Step 7: Write the Final Solution

Combine the results to write the final antiderivative:The indefinite integral of \( \frac{x+4}{x^{2}-16} \) is:\[\ln|x-4| + C\] where \( C \) is the constant of integration.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a handy technique used in integrating rational functions, like those seen in this type of problem. It's particularly useful when you have a fraction where the numerator is of a lower degree than the denominator. In our case, we're working with the fraction \( \frac{x+4}{x^2-16} \). The denominator is a difference of squares, giving us factors \((x-4)\) and \((x+4)\). Breaking this fraction into simpler parts makes integration straightforward.

• Express the fraction as a sum of simpler fractions: \( \frac{A}{x-4} + \frac{B}{x+4} \).
• Multiply both sides by the original denominator to eliminate the fraction: \( x+4 = A(x+4) + B(x-4) \).
• Expand and combine like terms to form an equation involving terms of \( x \) and constants.
• Solve for coefficients \( A \) and \( B \) using algebraic techniques such as equating coefficients and systems of equations. Here, \( A = 1 \) and \( B = 0 \).

By decomposing the fraction, integration becomes simpler as each component can be integrated separately.

Rational Functions

Rational functions are expressions formed by dividing two polynomials. They're common in calculus and integration due to their complexity and the variety of techniques involved in solving them.In the integration problem \( \int \frac{x+4}{x^2-16} \, dx \), the approach leverages the polynomial properties to simplify the process.

• The degree of the numerator (\( x+4 \), degree 1) is less than the degree of the denominator \( x^2-16 \) (degree 2), making it proper and suitable for partial fraction decomposition.
• The denominator is expressed as \((x-4)(x+4)\), simplifying the fraction into expressions that are easier to integrate.
• Rational functions like these are ideal candidates for partial fraction techniques, which break down complex expressions into manageable parts.

Understanding rational functions and their properties is essential, as it informs the choice of partial fraction decomposition in integral calculus.

Natural Logarithm

The natural logarithm \( \ln(x) \) frequently appears in solutions involving rational functions, especially when integrating simple fractions derived from partial fraction decomposition.When the decomposition gives us terms like \( \frac{1}{x-4} \), these can be directly integrated to natural logarithms. Here’s why:

• The integral \( \int \frac{1}{x-a} \, dx \) directly results in \( \ln|x-a| + C \), a standard formula for integrating fractions where the numerator is a constant and the denominator is linear.
• In our solution, this led us to \( \ln|x-4| + C \), where \( C \) is the constant of integration.
• The natural logarithm emerges due to the fundamental property: the derivative of \( \ln|x-a| \) is \( \frac{1}{x-a} \).

Always remember, when integrating and dealing with expressions that simplify to these forms, expect to see the natural logarithm as part of your result. It's a key component in many calculus problems!