Problem 32 - Precalculus

Question

Find the partial-fraction decomposition for each rational function. $$\frac{x^{2}}{\left(x^{2}+9\right)^{2}}$$

Step-by-Step Solution

Answer

The decomposition is \( \frac{1}{x^2 + 9} - \frac{9}{(x^2 + 9)^2} \).

Step 1: Identify the form of the rational function

The rational function given is \( \frac{x^{2}}{(x^{2} + 9)^{2}} \). This function has a quadratic expression \( x^{2} + 9 \) raised to the power of 2 as the denominator.

Step 2: Determine the decomposition setup

For the expression \( (x^2 + 9)^2 \), the partial fraction decomposition will require terms for each power of \( x^2 + 9 \) starting from 1 up to 2. Our decomposition will look like: \( \frac{Ax + B}{x^2 + 9} + \frac{Cx + D}{(x^2 + 9)^2} \), where \( A, B, C, \) and \( D \) are constants to be determined.

Step 3: Clear the denominator

Multiply both sides by \( (x^2 + 9)^2 \) to eliminate denominators. This results in: \( x^2 = (Ax + B)(x^2 + 9) + (Cx + D) \).

Step 4: Expand and combine like terms

Expand the right side: \((Ax + B)(x^2 + 9) = Ax^3 + 9Ax + Bx^2 + 9B\). Combine terms: \( Ax^3 + Bx^2 + 9Ax + 9B + Cx + D \).

Step 5: Set up and solve equations for coefficients

Compare coefficients with the left side equation \( x^2 = 0x^3 + x^2 + 0x + 0 \): 1. For \( x^3 \): \( A = 0 \) 2. For \( x^2 \): \( B = 1 \) 3. For \( x \): \( 9A + C = 0 \) 4. Constant term: \( 9B + D = 0 \)Solving, we have \( A = 0, B = 1, C = 0, D = -9 \).

Step 6: Write the partial-fraction decomposition

Substitute the values of \( A, B, C, \) and \( D \) back into the decomposition form: \( \frac{0 \cdot x + 1}{x^2 + 9} + \frac{0 \cdot x - 9}{(x^2 + 9)^2} = \frac{1}{x^2 + 9} - \frac{9}{(x^2 + 9)^2} \).

Key Concepts

Rational Functions

A rational function is essentially a fraction where both the numerator and denominator are polynomials. They are called 'rational' because they are ratios of polynomials. In our original exercise, the rational function is \( \frac{x^2}{(x^2 + 9)^2} \). Rational functions can become quite complex, but they have many interesting properties:

• The degree of the polynomial in the numerator is 2, as is indicated by \( x^2 \).
• The denominator is a polynomial \((x^2 + 9)\) squared, which means its degree is 4.
• The form of the fraction is crucial because it helps determine how we can decompose it, which is what partial fraction decomposition involves.

Understanding these characteristics is essential for simplifying expressions, solving integrals, or analyzing asymptotic behavior. Rational functions are used extensively in calculus, algebra, and other fields of mathematics for theory and practical applications.

Quadratic Expressions

A quadratic expression is a polynomial of degree 2, having the standard form \( ax^2 + bx + c \). In our problem's context, the expression \( x^2 + 9 \) is quadratic. It's a simple form where:

• \( a = 1 \), representing the coefficient of \( x^2 \)
• \( b = 0 \), since there's no \( x \) term
• \( c = 9 \), the constant term

Quadratic expressions can often be factored into the product of two linear expressions when a relationship, like factoring or difference of squares, applies.In partial fraction decomposition, quadratic expressions in the denominator may require special attention as they can appear either in their original form or raised to a higher power, and as such, influence the decomposition differently. Such expressions are important for determining the structure of the potential numerators, which often have the linear form \( Ax + B \) when handled in partial fractions.

Constant Coefficients

Constant coefficients are numbers that stand by themselves in an equation - they represent fixed values in polynomial expressions. In the given task, constants come into play when we determine the terms to be used in partial fraction decomposition.Consider this decomposition for example:\[ \frac{Ax + B}{x^2 + 9} + \frac{Cx + D}{(x^2 + 9)^2} \]Here, \( A, B, C, \) and \( D \) are constants that must be found to satisfy the terms of the decomposition based on the original rational function.In our problem:

• \( A = 0 \), which simplifies our expression.
• \( B = 1 \), indicating a direct contribution to the linear form.
• \( C = 0 \) equates the \( x \) component to zero for the second term.
• \( D = -9 \), and this negative value adjusts the constant part of the decomposition.

These constants are essential for reconstructing the composed terms of the partial fraction decomposition. They help break down the rational function into simpler components for analysis and integration.