Problem 3 - Thomas Calculus

Question

Expand the quotients in Exercises \(1-8\) by partial fractions. $$ \frac{x+4}{(x+1)^{2}} $$

Step-by-Step Solution

Answer

\( \frac{x+4}{(x+1)^2} = \frac{1}{x+1} + \frac{3}{(x+1)^2} \).

Step 1: Identify the Denominator

First, we look at the denominator of the given rational expression. Here, the denominator is \((x+1)^2\), which is a repeated linear factor.

Step 2: Set Up Partial Fraction Decomposition

Since \((x+1)^2\) is a repeated linear factor, the partial fraction decomposition will be of the form: \( \frac{x+4}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2} \).

Step 3: Clear the Denominator

Multiply both sides by \((x+1)^2\) to clear the denominator, which results in: \( x+4 = A(x+1) + B \).

Step 4: Expand and Simplify

Expand the right side: \( A(x+1) + B = Ax + A + B \). Thus, the equation becomes \( x+4 = Ax + A + B \).

Step 5: Equate Coefficients

Compare the coefficients of corresponding powers of \(x\) from both sides of the equation: for \(x\), we have \(1 = A\), and for the constant, we have \(4 = A + B\).

Step 6: Solve for Constants

From the equation \(1 = A\), we find that \(A = 1\). Substitute \(A = 1\) into \(4 = A + B\) to get \(4 = 1 + B\). Therefore, \(B = 3\).

Step 7: Write the Final Decomposition

Substitute the values of \(A\) and \(B\) back into the partial fraction decomposition to get: \( \frac{x+4}{(x+1)^2} = \frac{1}{x+1} + \frac{3}{(x+1)^2} \).

Key Concepts

Rational Expression

A rational expression, at its core, is a fraction where both the numerator and the denominator consist of polynomials. In the exercise, we have \( \frac{x+4}{(x+1)^{2}} \). This is a classic example of a rational expression.
Polynomials in the numerator and denominator define the complexity of the rational expression.
They often need simplification or even decomposing to make algebraic manipulations easier.

• Numerator: Here, it's a simple polynomial, \(x + 4\).
• Denominator: A repeated linear factor, \((x+1)^2\).

Recognizing these elements is the first step in performing partial fraction decomposition.

Repeated Linear Factor

A repeated linear factor occurs when a linear factor is raised to a power greater than one, such as \((x+1)^2\).
This requires special attention in partial fraction decomposition.
Why? Because each power of the linear factor up to the highest power in the denominator needs its own term in the decomposition.

• Basic linear factor: \(x+1\)
• Repeated factor: \((x+1)^2\)

Set up as: \(\frac{A}{x+1} + \frac{B}{(x+1)^2}\).
This ensures the decomposition is comprehensive. It accounts for all contributions from each power of the repeated factor.

Equate Coefficients

When the denominator is cleared, as seen in this example, the next step is to equate coefficients from both sides of the equation. This technique allows us to solve for the constants in our partial fraction decomposition.
The equation post-clearing is: \(x+4 = Ax + A + B\).

• Equate the coefficients of \(x\): Solves for \(A\).
• Equate the constants (no \(x\)): Solves for \(B\).

For this case:

• For the coefficient of \(x\): \(1 = A\), hence \(A = 1\).
• For the constant: \(4 = A + B\), substitute \(A = 1\), \(B = 3\).

Thus, equating coefficients effectively isolates each variable, leading us to the solution.

Clear the Denominator

In order to make the partial fraction decomposition more approachable, clearing the denominator is essential. This step simplifies the expression by multiplying through by the common denominator, thus removing fractions.
In our example:

• We multiply through by \((x+1)^2\): \(x+4 = A(x+1) + B\).

By doing so:

• We eliminate the complexity of fractions, facilitating easy comparison.
• It sets the stage for further operations like expanding and equating coefficients.

This clever algebraic trick simplifies the problem, making subsequent steps straightforward.