Problem 8 - Calculus Single Variable

Question

Write down the form of the partial fraction decomposition of the given rational function. Do not explicitly calculate the coefficients. \(\frac{3 x^{5}+x+1}{x^{2}\left(x^{2}-1\right)^{2}}\)

Step-by-Step Solution

Answer

\[ \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2} + \frac{E}{x+1} + \frac{F}{(x+1)^2} \]

Step 1: Identify the Denominator

First, let's factor the denominator: \( x^2 (x^2 - 1)^2 \). Notice that \( x^2 - 1 \) can be further factored as \((x-1)(x+1)\). So, the denominator can be written as \( x^2 (x-1)^2 (x+1)^2 \).

Step 2: Set Up the Partial Fraction Decomposition

For the partial fraction decomposition of \( x^2 (x-1)^2 (x+1)^2 \), each factor will have its own term(s) in the decomposition. 1. For \( x^2 \), use: \( \frac{A}{x} + \frac{B}{x^2} \) 2. For \( (x-1)^2 \), use: \( \frac{C}{x-1} + \frac{D}{(x-1)^2} \) 3. For \( (x+1)^2 \), use: \( \frac{E}{x+1} + \frac{F}{(x+1)^2} \).

Step 3: Write the Partial Fraction Decomposition

Combine all terms from each factor: \[ \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2} + \frac{E}{x+1} + \frac{F}{(x+1)^2} \].

Key Concepts

Rational Functions

Rational functions are expressions in mathematics that involve fractions in which both the numerator and the denominator are polynomials. For example, the given function in the exercise is \( \frac{3x^5 + x + 1}{x^2(x^2 - 1)^2} \). Here, the numerator is a polynomial \( 3x^5 + x + 1 \), and the denominator is \( x^2(x^2 - 1)^2 \). Rational functions can often be complex due to the higher degree of polynomials involved.

Studying these functions is crucial because they model many real-world phenomena and are essential in calculus, algebra, and advanced mathematics. When handling rational functions, a primary goal often is to simplify or decompose them for better understanding and to facilitate further operations like integration. This is where partial fraction decomposition comes in handy.

Partial fraction decomposition allows breaking down complex rational expressions into simpler fractions that are easy to integrate or differentiate. This technique is incredibly useful when dealing with integrals involving rational functions, making it an important tool in calculus.

Denominator Factoring

Factoring the denominator of a rational function is an essential step in partial fraction decomposition. In this exercise, the denominator \( x^2(x^2 - 1)^2 \) was carefully factored to assist in the decomposition process. Initially, notice how \( x^2 - 1 \) can be factored further into \((x-1)(x+1)\). Thus, the full factorization of the given denominator results in \( x^2(x-1)^2(x+1)^2 \).

This factorization allows us to recognize distinct components in the denominator :

• The polynomial \( x^2 \)
• The linear factors \( (x-1) \) and \( (x+1) \)

The purpose of this factorization is to identify all the distinct linear or quadratic factors that contribute to the decomposition process. With these factors, each can then be addressed separately by representing them with appropriate terms in the partial fraction decomposition.

Understanding how to factor efficiently helps simplify the decomposition and ensures that no step is overlooked in solving more complex functions.

Coefficients Identification

In partial fraction decomposition, identifying coefficients is the process of finding values for the variables that stand beside the fractions. Although in this exercise, explicit calculation of coefficients is not required, setting up the decomposition sets the stage for how these values could be determined.

Consider the setup:

• For the factor \( x^2 \), the terms include \( \frac{A}{x} \) and \( \frac{B}{x^2} \).
• For \( (x-1)^2 \), the terms are \( \frac{C}{x-1} \) and \( \frac{D}{(x-1)^2} \).
• For \( (x+1)^2 \), the terms appear as \( \frac{E}{x+1} \) and \( \frac{F}{(x+1)^2} \).

Each of these terms will have coefficients (A, B, C, D, E, F) that you need to solve for typically by equating the expression obtained from the partial fraction to the original rational function, multiplying through by the common denominator, and solving the resulting system of equations.

The core idea of coefficient identification is essential for fully completing the decomposition. It allows one to express a complex rational function into simpler, easily workable terms.