Problem 21 - Biocalculus Calculus for the Life Sciences

Question

$$\begin{array}{c}{\text { If a factor of the denominator is an irreducible quadratic, }} \\ {\text { such as } x^{2}+1, \text { the corresponding partial fraction has a }} \\ {\text { linear numerator. For instance, }} \\\ {f(x)=\frac{2 x^{2}+x+1}{x\left(x^{2}+1\right)}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}}\end{array}$$ $$\begin{array}{l}{\text { Determine the values of } A, B \text { , and } C \text { and use them to }} \\ {\text { evaluate } f f(x) d x .}\end{array}$$

Step-by-Step Solution

Answer

The values are \( A = 0 \), \( B = 2 \), and \( C = 1 \).

Step 1: Set up the equation for partial fractions

The function is given as \( f(x) = \frac{2x^2 + x + 1}{x(x^2 + 1)} \). We express it in terms of partial fractions: \( \frac{A}{x} + \frac{Bx + C}{x^2 + 1} \).

Step 2: Combine the partial fractions

Write the equation \( \frac{A}{x} + \frac{Bx + C}{x^2 + 1} = \frac{Ax^2 + A(Bx + C)}{x(x^2 + 1)} \) to form a common denominator and combine the terms.

Step 3: Simplify and equate numerators

Simplify the expression in the numerator to get \( Ax^2 + (Bx + C)x + A = Ax^2 + Bx^2 + Cx + A \). The original numerator is \( 2x^2 + x + 1 \), leading us to the equation \( A + Bx^2 + Cx = 2x^2 + x + 1 \).

Step 4: Match coefficients to solve for A, B, and C

By comparing the coefficients of corresponding powers of \( x \), we have: \( A = 0 \), \( B = 2 \), and \( C = 1 \).

Step 5: Verify the solution

Substitute \( A = 0 \), \( B = 2 \), and \( C = 1 \) into the partial fractions equation and simplify the expression to ensure it matches the original function. This verifies our solution is correct.

Key Concepts

Irreducible Quadratic

In algebra, an irreducible quadratic is a quadratic expression that cannot be factored into linear factors with real coefficients. For example, the quadratic expression \(x^2 + 1\) is considered irreducible over the real numbers because it has no real roots. In the context of partial fractions, dealing with irreducible quadratics is crucial because these components of the denominator cannot be broken down into simpler linear terms.

When you encounter an irreducible quadratic in the denominator, such as \(x^2 + 1\) in the original exercise, it suggests a more complex structure for the corresponding partial fraction. Here, you would expect the numerator over this irreducible quadratic to be a linear equation, typically in the form \(Bx + C\). This is different from having just a constant numerator when you deal with simple linear factors.

Linear Numerator

The concept of a linear numerator comes into play specifically while dealing with partial fractions, especially when working with irreducible quadratics in the denominator. Instead of merely having a single constant value, the numerator takes the form of \(Bx + C\), which is linear in nature.

This form matches the degree of the numerator to the degree of the irreducible quadratic denominator minus one. This helps in ensuring that the fraction remains a proper one, which is essential for finding its partial fraction decomposition. In our exercise, given the denominator \(x^2 + 1\), the linear numerator is expressed as \(Bx + C\) over \(x^2 + 1\).

Including variables in the numerator allows us to cover all terms in the original expression, providing flexibility for matching coefficients as explained later. This format simplifies subsequent algebraic manipulations and is a crucial part of solving the partial fractions efficiently.

Coefficient Matching

Coefficient matching is a technique used in the process of decomposing complex fractions into partial fractions. The goal is to equate the coefficients of the numerator's terms after expressing them in a common denominator to those of the original function's numerator. This allows the determination of unknown constants in the numerators of the partial fractions.

In the original exercise, after setting up the partial fraction decomposition, we end up with both sides having a polynomial in the numerator. Here, the task becomes efficiently matching coefficients of the powers of \(x\). The equation formed for matching in this case was \(A + Bx^2 + Cx = 2x^2 + x + 1\), where each coefficient of powers of \(x\) from both sides is compared.

• For \(x^2\), equate: \(A + B = 2\)
• For \(x\), equate: \( C = 1\)
• Constant term, equate: \(A = 0\)

Step by step, solving these simultaneous equations leads you to find the values of \(A\), \(B\), and \(C\). This process not only solidifies understanding but ensures the integrity of the partial fractions you've formed. In our case, it resulted in \(A = 0\), \(B = 2\), and \(C = 1\), beautifully verifying the correctness of our decomposition.