Problem 19 - Biocalculus Calculus for the Life Sciences

Question

If a linear factor in the denominator of a rational function is repeated, there will be two corresponding partial fractions. For instance, $$f(x)=\frac{5 x^{2}+3 x-2}{x^{2}(x+2)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+2}$$ $$\begin{array}{l}{\text { Determine the values of } A, B, \text { and } C \text { and use them to }} \\ {\text { evaluate } \int f(x) d x .}\end{array}$$

Step-by-Step Solution

Answer

\( A = 2, B = -1, C = 3 \), and \( \int f(x) \, dx = 2\ln|x| + \frac{1}{x} + 3\ln|x+2| + C \).

Step 1: Set Up Partial Fractions

Given the rational function \( f(x) = \frac{5x^2 + 3x - 2}{x^2(x+2)} \), we need to find constants \( A, B, \) and \( C \) such that \( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+2} \) equals the original function after combining the fractions. The goal is to express \( \frac{5x^2 + 3x - 2}{x^2(x+2)} \) as \( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+2} \).

Step 2: Clear the Denominator

Multiply through by \( x^2(x+2) \) to eliminate the denominators: \[ 5x^2 + 3x - 2 = A(x)(x+2) + B(x+2) + C(x^2). \] Expand and simplify the right-hand side.

Step 3: Expand and Simplify

Expand the expressions:- \( A(x)(x+2) = Ax^2 + 2Ax \)- \( B(x+2) = Bx + 2B \)- \( C(x^2) = Cx^2 \)This results in:\[ (A + C)x^2 + (2A + B)x + 2B. \]

Step 4: Equate Coefficients

By comparing coefficients with the left-hand side \( 5x^2 + 3x - 2 \), we get the following equations:- For \( x^2 \): \( A + C = 5 \)- For \( x \): \( 2A + B = 3 \)- Constant: \( 2B = -2 \).

Step 5: Solve the System of Equations

Solve for \( B \):\[ 2B = -2 \quad \Rightarrow \quad B = -1. \] Substitute \( B = -1 \) in \( 2A + B = 3 \):\( 2A - 1 = 3 \quad \Rightarrow \quad 2A = 4 \) hence \( A = 2 \).Substitute \( A = 2 \) in \( A + C = 5 \):\( 2 + C = 5 \quad \Rightarrow \quad C = 3. \)

Step 6: Write Partial Fraction Decomposition

Now that we have \( A = 2 \), \( B = -1 \), and \( C = 3 \), the partial fraction decomposition is:\[ \frac{5x^2 + 3x - 2}{x^2(x+2)} = \frac{2}{x} - \frac{1}{x^2} + \frac{3}{x+2}. \]

Step 7: Evaluate the Integral

Integrate each term separately:\[ \int \left( \frac{2}{x} \right) dx = 2\ln|x| + C_1, \]\[ \int \left( -\frac{1}{x^2} \right) dx = \frac{1}{x} + C_2, \]\[ \int \left( \frac{3}{x+2} \right) dx = 3\ln|x+2| + C_3. \]Combine these results to write the full integral:\[ \int f(x) \, dx = 2\ln|x| + \frac{1}{x} + 3\ln|x+2| + C. \]

Key Concepts

Rational Functions

Rational functions are a type of function expressed as the ratio of two polynomials. The general form is \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \). These functions can have vertical asymptotes, horizontal asymptotes, or slant asymptotes depending on the degrees of the polynomial in the numerator and the denominator.

• If the degree of \( P(x) \) is less than the degree of \( Q(x) \), then the horizontal asymptote is \( y = 0 \).
• If the degree of \( P(x) \) is equal to the degree of \( Q(x) \), the horizontal asymptote is \( y = \frac{a}{b} \) where \( a \) and \( b \) are the leading coefficients of \( P(x) \) and \( Q(x) \) respectively.
• If the degree of \( P(x) \) is greater than the degree of \( Q(x) \), a slant asymptote may occur.

Partial fraction decomposition is a tool that simplifies complex rational functions into simpler fractions, facilitating easier integration or solving of equations.

Integration

Integration is a fundamental concept in calculus, helping to find areas, volumes, and even solve differential equations. When dealing with rational functions, integration can become complex due to their form. However, by using partial fraction decomposition, we can transform complex rational functions into simpler forms that are easier to integrate.
Let's look at our example: the function \(- \frac{1}{x^2}\) becomes \(-x^{-2}\), for which the integral follows the power rule: \(- \int x^{-2} \, dx = \frac{1}{x} + C \).

• The integral of \( \frac{2}{x} \) is straightforward, resulting in \( 2\ln|x| \).
• Similarly, \( \frac{3}{x+2} \) integrates to \( 3\ln|x+2| \).

This simplification makes the integration process manageable and straightforward.

Systems of Equations

To decompose a rational function into partial fractions, we often need to solve systems of equations. This is necessary when matching coefficients after expressing the original polynomial in simpler terms.
In our exercise, we found the equations:

• \( A + C = 5 \)
• \( 2A + B = 3 \)
• \( 2B = -2 \)

Solving these equations helped us determine the values for \( A, B, \) and \( C \).

• We solved for \( B \) first, leading to \( B = -1 \).
• Substituting \( B \) into the second equation yielded \( A = 2 \).
• Finally, substituting \( A = 2 \) into the first equation provided \( C = 3 \).

Solving these linear equations is crucial for the decomposition and helps students understand dependencies between different parts of a polynomial equation.

Polynomial Division

Polynomial division is similar to long division with numbers, used when dividing one polynomial by another. Before doing partial fraction decomposition, it's important to check if the degree of the numerator is greater than or equal to the degree of the denominator. If it is, polynomial division is required.
In this problem, the numerator \( 5x^2 + 3x - 2 \) doesn't have a higher degree than the denominator \( x^2(x+2) \), so no division was needed.

• Polynomial division helps simplify rational functions to a proper form.
• A proper rational function has a numerator degree less than the denominator.

Understanding polynomial division is essential for simplifying complex fractions, setting up partial fractions and ensuring an accurate decomposition process.