Problem 12 - Precalculus: Mathematical for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{2 x}{(x-1)(x+1)}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{1}{x-1} + \frac{1}{x+1} \).

Step 1: Understand the Rational Function

We are given the rational function \( \frac{2x}{(x-1)(x+1)} \). Our goal is to express this function as a sum of simpler fractions, each called a partial fraction.

Step 2: Setup for Partial Fractions

The denominator \((x-1)(x+1)\) consists of two distinct linear factors. Thus, the decomposition will have the form: \( \frac{A}{x-1} + \frac{B}{x+1} \), where \(A\) and \(B\) are constants to be determined.

Step 3: Combine the Fractions

First, rewrite the expression \( \frac{A}{x-1} + \frac{B}{x+1} \) with a common denominator: \[ \frac{A(x+1) + B(x-1)}{(x-1)(x+1)} \].

Step 4: Match the Numerators

Equate the numerators from the original function and the combined partial fractions:\[ 2x = A(x+1) + B(x-1) \].Expand and simplify to get:\[ 2x = (A+B)x + (A-B) \].

Step 5: Determine Unknown Constants

By comparing coefficients, we form the system of equations:- \( A + B = 2 \) (coefficient of \(x\)),- \( A - B = 0 \) (constant term).By solving these equations:1. \( A - B = 0 \) implies \( A = B \).2. Substitute \( A = B \) into \( A + B = 2 \), we get \( 2A = 2 \) leading to \( A = 1 \) and \( B = 1 \).

Step 6: Write the Final Decomposition

Now that we have \( A = 1 \) and \( B = 1 \), substitute these back into the partial fraction expression:\[ \frac{2x}{(x-1)(x+1)} = \frac{1}{x-1} + \frac{1}{x+1} \].

Key Concepts

Understanding Rational Functions

A rational function is a quotient of two polynomials. In our case, the given function is \( \frac{2x}{(x-1)(x+1)} \), where the numerator is \(2x\) and the denominator is \((x-1)(x+1)\). Rational functions can be complex, so breaking them into simpler parts, called partial fractions, helps in analysis and integration. In this context, the task involves expressing the given rational function as a sum of simpler, separate fractions. This approach is particularly useful in calculus when integrating complex rational functions. Overall, the decomposition process makes handling rational functions much easier and more intuitive.

Forming and Solving Systems of Equations

To find the values of the constants in the partial fraction decomposition, we create and solve a set of equations. These equations arise when we equate coefficients from matching numerators of the original function and the decomposed sum. For the function \( \frac{2x}{(x-1)(x+1)} \), this results in comparing:

• \( 2x = (A + B)x + (A - B) \)

By matching coefficients:

• For \(x\): \( A + B = 2 \)
• For constant term: \( A - B = 0 \)

Solving these equations simultaneously:

• From \( A - B = 0 \), we find \( A = B \).
• Substituting \( A = B \) into \( A + B = 2 \) results in \( 2A = 2 \), leading to \( A = 1 \) and \( B = 1 \).

This system of equations is a crucial step in identifying the constants for your partial fractions and shows how algebraic techniques interlink with calculus.

The Role of Linear Factors

When we decompose a rational function, recognizing linear factors is vital. Linear factors are simply factors in the polynomial of degree one, like \(x - 1\) or \(x + 1\), as seen in our denominator. These linear factors determine how we set up the partial fractions. For each linear factor in the denominator, we set a separate fraction:

• \( \frac{A}{x-1} \)
• \( \frac{B}{x+1} \)

Each of these fractions will have a constant in the numerator (like \(A\) and \(B\)).The decomposition process relies on these linear factors being distinct. If there were repeated factors, the setup would include additional terms, accounting for the multiplicities. Identifying these factors is the first step in the decomposition and helps in systematically solving for unknowns like \(A\) and \(B\). This method makes rational functions more manageable and is a powerful tool in mathematical problem solving.