Question

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

Short answer

\(\frac{1}{x} - \frac{x}{x^2 + 1}\)

Step-by-step solution

Recognize the Form of the Denominator

First, notice that the denominator is a product of two distinct factors: one linear factor, \(x\), and one irreducible quadratic factor, \(x^2 + 1\).

Set Up the Partial Fractions

Since the denominator is a product of a linear term and an irreducible quadratic term, set up the partial fraction decomposition as follows: \[\frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}\]

Combine the Fractions on the Right Side

Combine the fractions on the right side over a common denominator, which is \(x(x^2 + 1)\): \[\frac{A(x^2 + 1) + (Bx + C)x}{x(x^2 + 1)}\]

Equate Numerators

Since the denominators are the same, equate the numerators: \[1 = A(x^2 + 1) + (Bx + C)x\]

Expand and Collect Like Terms

Expand the right-hand side and collect like terms: \[1 = Ax^2 + A + Bx^2 + Cx\] Combine like terms: \[1 = (A + B)x^2 + Cx + A\]

Form System of Equations

Match the coefficients of corresponding terms to form a system of equations: 1. For the coefficient of \(x^2\): \(A + B = 0\) 2. For the coefficient of \(x\): \(C = 0\) 3. For the constant term: \(A = 1\)

Solve the System of Equations

Solve the equations: From equation 3: \(A = 1\) Substitute \(A = 1\) into equation 1: \(1 + B = 0\) Solve for \(B\): \(B = -1\) From equation 2: \(C = 0\)

Write the Partial Fraction Decomposition

Substitute \(A\), \(B\), and \(C\) back into the partial fractions: \[\frac{1}{x\left(x^2 + 1\right)} = \frac{1}{x} + \frac{-x}{x^2 + 1}\]

Simplify

Simplify the expression: \[\frac{1}{x} - \frac{x}{x^2 + 1}\]

Key concepts

rational expressions

When dealing with partial fraction decomposition, we often start with rational expressions. A rational expression is simply a fraction where both the numerator and the denominator are polynomials. These expressions can often be complex, and breaking them into simpler parts using partial fraction decomposition makes them easier to work with. In our exercise, we have the rational expression: \[ \frac{1}{x(x^2 + 1)} \] Understanding the interaction between the numerator and denominator polynomials is key to proceeding with the decomposition.

irreducible quadratic factor

In our expression, we noted a critical feature: an irreducible quadratic factor. There are different kinds of factors that we deal with in partial fraction decomposition. Linear factors (like \( x \)) are straightforward. However, factors like \( x^2 + 1 \) are quadratics that can't be factored further over the real numbers. We call these irreducible. The presence of an irreducible quadratic factor changes how we set up our partial fractions: \[ \frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} \] Because of the quadratic term, the numerator of our partial fraction for this part must also be a linear polynomial (\( Bx + C \)).

system of equations

To find the values of \( A \), \( B \), and \( C \), we set up a system of equations. This comes from equating the numerators once we combine the partial fractions on a common denominator. Here’s the step where we equate numerators: \[ 1 = A(x^2 + 1) + (Bx + C)x \] Expanding this gives us: \[ 1 = Ax^2 + A + Bx^2 + Cx \] Combining like terms, we have: \[ 1 = (A + B)x^2 + Cx + A \] Now, we form a system of equations by looking at the coefficients of \( x^2 \), \( x \), and the constant term: \- Coefficient of \( x^2 \): \( A + B = 0 \)\- Coefficient of \( x \): \( C = 0 \)\- Constant term: \( A = 1 \) This system can now be solved step by step to find the values of \( A \), \( B \), and \( C \).

linear factor

Linear factors are simpler parts of our rational expressions. In our exercise, \( x \) is a linear factor. These are first-degree polynomials, and their partial fraction decomposition is straightforward: \[ \frac{A}{x} \] The presence of a linear factor helps to simplify our work because its partial fraction component only involves a single term. Understanding this makes it easier when learning how to handle more complex terms like irreducible quadratics in decomposition.

numerator and denominator

Successful partial fraction decomposition requires careful handling of both the numerator and the denominator.\br The numerator is the polynomial on top (in our case, just \( 1 \)) and the denominator is the product of linear and quadratic factors.\br By breaking it down as we did: \[ \frac{1}{x(x^2 + 1)} = \frac{1}{x} + \frac{-x}{x^2 + 1} \] We see how the decomposed terms have numerators (\( 1 \) and \( -x \)) that exactly match the necessary form for our factors.\br The careful balance of these components ensures that our decomposition is accurate and useful for further applications.