Problem 24 - A Graphical Approach to College Algebra

Question

Find the partial fraction decomposition for each rational expression. $$\frac{x^{4}+1}{x\left(x^{2}+1\right)^{2}}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{1}{x} - \frac{2x}{(x^2+1)^2} \).

Step 1: Identify the Denominator

The given rational expression is \( \frac{x^{4}+1}{x(x^{2}+1)^{2}} \). To start, identify the factors in the denominator, which are \( x \) and \( (x^{2}+1)^{2} \).

Step 2: Set Up the Partial Fraction Decomposition

Since the denominator has a linear factor \( x \) and a repeated irreducible quadratic factor \( (x^{2}+1)^{2} \), the partial fraction decomposition will be: \[ \frac{A}{x} + \frac{Bx+C}{x^{2}+1} + \frac{Dx+E}{(x^{2}+1)^{2}}. \]

Step 3: Combine and Clear the Denominator

Multiply both sides of the equation by the denominator \( x(x^{2}+1)^{2} \) to eliminate the fraction: \[ x^{4}+1 = A(x^{2}+1)^{2} + (Bx+C)x(x^{2}+1) + (Dx+E)x. \]

Step 4: Expand and Collect Like Terms

Expand the right hand side to express it in terms of powers of \( x \):\[ A(x^{4} + 2x^{2} + 1) + (Bx^3 + Bx + Cx^{2} + C) + (Dx^2 + Ex) = x^4 + 1. \] Collect like terms for \( x^4, x^3, x^2, x^1, \) and the constant term.

Step 5: Solve the System of Equations

Equating coefficients from both sides, solve the following system: 1. \( A = 1 \) from \( x^4 \) terms. 2. \( B = 0 \) from \( x^3 \) terms. 3. \( 2A + C + D = 0 \) from \( x^2 \) terms. 4. \( B + E = 0 \) from \( x^1 \) terms. 5. \( A + C = 1 \) for constant term.

Step 6: Substitute and Solve for Coefficients

From \( A=1 \) and \( B=0 \), we have: - Substitute \( A = 1 \) in 3 and 5 to get: 1. \( 2 + C + D = 0 \) 2. \( 1 + C = 1 \) - Solving the system, we get \( C=0, D=-2 \), and \( E=0 \).

Step 7: Write the Resulting Decomposition

The partial fraction decomposition is: \[ \frac{1}{x} + \frac{0 \cdot x + 0}{x^2+1} + \frac{-2 \cdot x + 0}{(x^2+1)^2}. \] Simplified, this is: \[ \frac{1}{x} - \frac{2x}{(x^2+1)^2}. \]

Key Concepts

Rational Expression

When we talk about a rational expression, we refer to a fraction that involves polynomials in both its numerator and denominator. In this exercise, we are asked to decompose a rational expression into simpler parts using a technique known as partial fraction decomposition. The original rational expression here is \( \frac{x^{4}+1}{x(x^{2}+1)^{2}} \). This expression consists of a polynomial numerator \( x^4 + 1 \) and a complex polynomial denominator with multiple distinct and repeated factors. Rational expressions are crucial in calculus and algebra because they can be manipulated, simplified, or integrated more easily when decomposed into simpler, separate fractions.

Denominator Factors

Understanding the factors of the denominator is vital in performing partial fraction decomposition. In the given expression \( \frac{x^{4} + 1}{x(x^{2} + 1)^{2}} \), the denominator can be factored into \( x \) and \( (x^2 + 1)^2 \). The first factor, \( x \), is a linear factor, while \( (x^2 + 1) \) is an irreducible quadratic factor that is repeated. These components dictate the structure of the partial fraction decomposition. The repeated quadratic factor requires breaking the expression into two separate fractions with unknown coefficients placed as numerators. Recognizing these factors sets the stage for setting up and solving the initial decomposition form.

System of Equations

Once the rational expression's denominator has been factorized, the next step involves forming a system of equations. By expressing the rational function as a sum of its partial fractions, and then multiplying through by the common denominator to clear fractions, we'll arrive at an equality of polynomials. Here, it becomes crucial to expand the expression on the right-hand side and compare coefficients of the corresponding powers of \( x \) from both sides of the equation. The system of equations established from this process features equations directly linking the coefficients of the expanded terms, thus allowing us to solve for the unknown coefficients, like \( A \), \( B \), \( C \), \( D \), and \( E \) in this particular exercise.

Coefficient Comparison

Coefficient comparison is the heart of solving the system of equations from the partial fraction setup. Once we've expressed both sides of the equation as polynomials of the same degree, we equate each corresponding coefficient from each side. This process is applied to each term: constants, as well as coefficients of \( x \), \( x^2 \), \. and onwards. Each term yields one equation in the system. For our exercise, the key equations extracted were: \( A = 1 \), \( B = 0 \), and further equations leading to \( C = 0 \), \( D = -2 \), and \( E = 0 \). Solving these equations gives the coefficients for the partial fraction decomposition, simplifying our efforts to understand the expression deeply.