Find the partial fraction decomposition for each rational expression. See Examples 1–4. (3x - 1)/(x(2x² + 1)²)

Verified step by step guidance

1. Identify the form of the denominator: it is composed of a linear factor \(x\) and a repeated irreducible quadratic factor \((2x^2 + 1)^2\).
2. Set up the partial fraction decomposition with unknown constants. Since \(x\) is linear, assign a constant numerator \(A\) over \(x\). For the repeated quadratic \((2x^2 + 1)^2\), assign linear numerators \(Bx + C\) over \((2x^2 + 1)\) and \(Dx + E\) over \((2x^2 + 1)^2\). So, write: \(\frac{3x - 1}{x(2x^2 + 1)^2} = \frac{A}{x} + \frac{Bx + C}{2x^2 + 1} + \frac{Dx + E}{(2x^2 + 1)^2}\)
3. Multiply both sides of the equation by the common denominator \(x(2x^2 + 1)^2\) to clear the fractions. This gives: \$3x - 1 = A(2x^2 + 1)^2 + (Bx + C) x (2x^2 + 1) + (Dx + E) x$
4. Expand the right-hand side completely by first expanding \((2x^2 + 1)^2\), then distributing \(A\), \((Bx + C) x\), and \((Dx + E) x\) terms. Collect like terms of powers of \(x\) (e.g., \(x^4\), \(x^3\), \(x^2\), \(x\), and constants).
5. Equate the coefficients of corresponding powers of \(x\) on both sides of the equation to form a system of linear equations in terms of \(A\), \(B\), \(C\), \(D\), and \(E\). Solve this system to find the values of these constants.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a complex rational expression as a sum of simpler fractions. This technique is especially useful for integrating rational functions or simplifying expressions. It involves breaking down the denominator into factors and assigning unknown constants to each term to solve for.

Factorization of Polynomials

Understanding how to factor polynomials is essential for partial fraction decomposition. The denominator must be factored into linear and/or irreducible quadratic factors, including repeated factors. Recognizing these factors helps determine the form of the partial fractions needed for the decomposition.

Handling Repeated Quadratic Factors

When the denominator contains repeated irreducible quadratic factors, each power of the factor must be included in the decomposition with its own numerator. For example, for (2x^2 + 1)^2, terms with denominators (2x^2 + 1) and (2x^2 + 1)^2 appear, each with a linear numerator. This ensures the decomposition accounts for all multiplicities.