Problem 8 - Calculus: One and Several Variables

Question

Decompose into partial fractions.$$\frac{1}{x\left(x^{2}+1\right)^{2}}$$.

Step-by-Step Solution

Answer

To decompose the rational function \(\frac{1}{x(x^2 + 1)^2}\) into partial fractions, we first identify the form of the partial fractions decomposition, which is: \(\frac{A}{x} + \frac{Bx + C}{x^2 + 1} + \frac{Dx + E}{(x^2 + 1)^2}\) Next, we clear the denominators and find the coefficients A, B, C, D, and E, which are: \(A = 1, \quad B = -\frac{2}{3}, \quad C = 0, \quad D = 6, \quad E = -8\) Thus, the partial fractions decomposition is: \(\frac{1}{x(x^2 + 1)^2} = \frac{1}{x} - \frac{\frac{2}{3}x}{x^2 + 1} + \frac{6x - 8}{(x^2 + 1)^2}\)

Step 1: Identify the form of the partial fractions

The denominator of the given fraction is $$x(x^2+1)^2$$. Here, we have three factors: One linear factor, $$x$$, and a quadratic factor, $$(x^2 +1)$$, with multiplicity 2. For each factor, we need to find corresponding constants in our partial fractions. The form of the partial fractions decomposition is: $$\frac{A}{x} + \frac{Bx+C}{x^2 + 1} + \frac{Dx+E}{(x^2 + 1)^2}$$

Step 2: Clear the denominators

Multiply both sides by the denominator $$x(x^2+1)^2$$ to eliminate the denominators: $$1 = A(x^2 + 1)^2 + (Bx + C)x(x^2 + 1) + (Dx + E)x$$

Step 3: Find the coefficients

Now, we need to find the constants A, B, C, D, and E. First, let's plug in values for x such that, some terms become zero. Setting x=0: $$1 = A(1)^2$$ Solving for A, we get: $$A = 1$$ Now, we need to find B, C, D, and E. We can achieve this by differentiating both sides with respect to x and substitute some suitable values. Let's differentiate the equation with respect to x: $$0 = 2(1)(x^2 + 1)(2x) + (Bx + C)(2x^2 + 2) + (Bx^2 + Cx)(1) + x(2x(Dx + E)) + (x^2 + 1)^2(2Dx + 2E)$$ Now, set x=0 again: $$0 = (C)(2)$$ Solving for C, we get: $$C = 0$$ Differentiate the above equation again to find B, D, and E: $$0 = 4(1)(x^2 + 1)(2) + 2(Bx^2 + Cx)(1) + B(2x) + 2x(2x(Dx + E)) + (2x^2 + 2)(2Dx + 2E)$$ Again, set x=0: $$0 = 4(2) + 0 + E(0)$$ Solving for E, we get: $$E = -8$$ Now, we still need to find B and D. We can plug in random values for x and solve the system of equations: Let x=1, then: $$1 = (1)(2 + 1)^2 + (B + 0)(1 + 2) + (D + (-8))(1)$$ Let x=-1, then: $$1 = (1)(2 + 1)^2 + ((-B) + 0)(1 + 2) + ((-D) + (-8))(-1)$$ Solving the system of equations, we get: $$B = -\frac{2}{3}, \quad D = 6$$

Step 4: Write the partial fractions decomposition

Now that we have the constants A, B, C, D, and E, we can write the partial fractions decomposition: $$\frac{1}{x(x^2 + 1)^2} = \frac{1}{x} - \frac{\frac{2}{3}x}{x^2 + 1} + \frac{6x - 8}{(x^2 + 1)^2}$$

Key Concepts

Linear Factor

In the context of partial fractions decomposition, a linear factor is a term in the denominator of the fraction that is first degree, meaning it can be expressed in the form of \( (ax + b) \). A simple example of a linear factor is \( x \), as it is in the first degree of the variable \( x \). In the given problem, \( x \) is identified as a linear factor in the denominator \( x(x^2+1)^2 \). This factor will correspond to a term in the partial fraction decomposition that has a constant numerator:

• For each linear factor, there is a term in the decomposition like \( \frac{A}{x} \), where \( A \) is a constant.

By recognizing a linear factor, we can systematically set up our partial fractions and solve for the constants using various methods such as substituting convenient values into the system of equations. This approach simplifies the process of breaking down a complex rational expression into simpler parts.

Quadratic Factor

In partial fractions, a quadratic factor is a polynomial in the denominator of degree two. In the exercise provided, \((x^2 + 1)\) is a quadratic factor. Quadratic factors can appear in a more complex form, especially if they have multiplicities higher than one, as in the case of \((x^2 + 1)^2\).

• For each unique quadratic factor of degree two, a term in the partial fractions decomposition looks like \( \frac{Bx + C}{x^2 + 1} \).
• If the quadratic factor is repeated like \((x^2 + 1)^2\), you must account for each occurrence with terms such as \( \frac{Dx + E}{(x^2 + 1)^2} \).

Quadratic factors complicate the partial fractions because they require finding two constants, \( B \) and \( C \), for each factor term. By expressing each repeated quadratic factor separately, handling these becomes manageable, allowing us to solve for the constants methodically.

Denominator

The denominator in a rational function is crucial for determining the form of partial fraction decomposition. It impacts how the expression is broken down into simpler fractions. In our example, the denominator is \(x(x^2+1)^2\), consisting of both linear and quadratic factors.

• The form of the denominator directly influences our approach to decomposition. Each distinct type of factor (linear or quadratic) contributes different terms to the decomposition.
• Understanding the structure of the denominator is essential because each factor must be accounted for separately when writing the form of the partial fractions.

By analyzing the denominator thoroughly, we can set up a proper structure for finding coefficients for each term in the partial fractions, making sure that every factor is considered in the breakdown.

Differentiation

Differentiation plays a key role in finding the unknown coefficients in partial fraction decomposition when simple substitution doesn't yield all the necessary values. It involves taking the derivative of both sides of the equation obtained after clearing denominators:

• Differentiation helps in generating additional equations that can be used to solve for unknowns like \( B \), \( C \), \( D \), and \( E \).
• By differentiating and setting specific values for \( x \), we can zero out certain terms, simplifying the solving process.

This technique is particularly useful in scenarios where plugging in typical values of \( x \) does not suffice to clear terms and isolate coefficients. As seen in the example, differentiating and solving the derivatives provides a pathway to determine the values of constants systematically.