Partial Fraction Decomposition Of (x^2+1)/((x^2-1)^2) A Comprehensive Guide

Introduction to Partial Fraction Decomposition

In the realm of algebraic manipulations, partial fraction decomposition stands out as a powerful technique to simplify rational functions. Specifically, we can express a complex rational function as a sum of simpler fractions. This method is invaluable in various areas of mathematics, including calculus (for integration) and engineering (for system analysis). In this article, we delve deep into the process of decomposing the given rational function, $(x^2+1)/((x^2-1)^2)$, into its partial fractions, providing a comprehensive step-by-step guide. Understanding partial fraction decomposition is not only crucial for solving mathematical problems but also enhances analytical skills in broader scientific contexts. The beauty of this method lies in its ability to transform complicated expressions into manageable parts, facilitating further analysis and computations. As we navigate through the intricacies of this decomposition, we will uncover the underlying principles and techniques that make this process both effective and elegant.

Understanding the Given Rational Function

Before diving into the decomposition process, let's closely examine the rational function at hand: $(x^2+1)/((x^2-1)^2)$. The key to partial fraction decomposition lies in understanding the structure of the denominator. Here, we observe that the denominator, $(x^2-1)^2$, can be further factored. Recall the difference of squares factorization: $a^2 - b^2 = (a - b)(a + b)$. Applying this to our denominator, we get $(x^2 - 1) = (x - 1)(x + 1)$. Thus, the denominator can be rewritten as $((x - 1)(x + 1))^2$, which simplifies to $(x - 1)^2(x + 1)^2$. This factorization is crucial because it reveals the repeated linear factors in the denominator. Repeated factors necessitate a specific approach in partial fraction decomposition, where each factor is considered with powers up to its multiplicity. Recognizing these factors and their multiplicities is paramount for setting up the correct form of the partial fractions. By understanding the foundational structure of the given rational function, we pave the way for a systematic decomposition process that ensures accurate and efficient results. This preliminary analysis is not merely a formality but a vital step in unraveling the complexity of the rational function and setting the stage for its simplification.

Setting Up the Partial Fraction Decomposition

The heart of partial fraction decomposition lies in setting up the correct form of the partial fractions. Given the factored denominator $(x - 1)^2(x + 1)^2$, we recognize that we have repeated linear factors. For each repeated linear factor, such as $(x - 1)^2$, we include terms with denominators ranging from the first power up to the power of repetition. Therefore, for $(x - 1)^2$, we include terms $A/(x - 1)$ and $B/(x - 1)^2$, where A and B are constants to be determined. Similarly, for $(x + 1)^2$, we include terms $C/(x + 1)$ and $D/(x + 1)^2$, where C and D are constants. Consequently, the partial fraction decomposition of $(x^2 + 1)/((x^2 - 1)^2)$ takes the form:

$(x^2 + 1)/((x^2 - 1)^2) = A/(x - 1) + B/(x - 1)^2 + C/(x + 1) + D/(x + 1)^2$

This equation sets the stage for solving for the unknowns A, B, C, and D. The next step involves clearing the denominators by multiplying both sides of the equation by the original denominator, $(x - 1)^2(x + 1)^2$. This process transforms the equation into a polynomial equation, which we can then use to solve for the constants. The careful setup of these partial fractions is the cornerstone of the entire decomposition process, ensuring that we accurately represent the original rational function as a sum of simpler fractions. By meticulously accounting for each factor and its multiplicity, we establish a solid foundation for the subsequent steps in the decomposition.

Clearing Denominators and Forming the Equation

To solve for the unknown constants A, B, C, and D in our partial fraction decomposition, we must first clear the denominators. This crucial step involves multiplying both sides of the equation:

$(x^2 + 1)/((x - 1)^2(x + 1)^2) = A/(x - 1) + B/(x - 1)^2 + C/(x + 1) + D/(x + 1)^2$

by the common denominator, which is $(x - 1)^2(x + 1)^2$. When we perform this multiplication, each fraction on the right-hand side is multiplied by the common denominator, effectively canceling out its own denominator. This process yields:

$x^2 + 1 = A(x - 1)(x + 1)^2 + B(x + 1)^2 + C(x + 1)(x - 1)^2 + D(x - 1)^2$

This equation is a polynomial equation, and it forms the basis for solving for the constants. Expanding the terms on the right-hand side will give us a polynomial in terms of x, with coefficients involving A, B, C, and D. This expansion is a critical step, as it transforms the equation into a form where we can equate coefficients of like powers of x. By carefully clearing the denominators, we eliminate the fractional form and transition to a more manageable polynomial equation. This equation now holds the key to unlocking the values of our unknown constants, allowing us to complete the partial fraction decomposition. The next step involves either substituting specific values of x or equating coefficients to solve for A, B, C, and D.

Solving for the Constants

With the equation $x^2 + 1 = A(x - 1)(x + 1)^2 + B(x + 1)^2 + C(x + 1)(x - 1)^2 + D(x - 1)^2$ in hand, we now embark on the task of solving for the constants A, B, C, and D. There are two primary methods to achieve this: the substitution method and the method of equating coefficients.

Substitution Method

The substitution method involves strategically choosing values of x that simplify the equation, ideally by making some terms zero. For instance, substituting $x = 1$ eliminates the terms with factors $(x - 1)$, leaving us with:

$1^2 + 1 = B(1 + 1)^2$

$2 = 4B$

$B = 1/2$

Similarly, substituting $x = -1$ eliminates the terms with factors $(x + 1)$, giving us:

$(-1)^2 + 1 = D(-1 - 1)^2$

$2 = 4D$

$D = 1/2$

Method of Equating Coefficients

The method of equating coefficients involves expanding the polynomial on the right-hand side, grouping like terms, and then equating the coefficients of corresponding powers of x on both sides of the equation. Expanding the right-hand side, we get:

$x^2 + 1 = A(x^3 + x^2 - x - 1) + B(x^2 + 2x + 1) + C(x^3 - x^2 - x + 1) + D(x^2 - 2x + 1)$

Grouping like terms, we have:

$x^2 + 1 = (A + C)x^3 + (A + B - C + D)x^2 + (-A + 2B - C - 2D)x + (-A + B + C + D)$

Now, we equate the coefficients of like powers of x:

• Coefficient of $x^3$: $A + C = 0$
• Coefficient of $x^2$: $A + B - C + D = 1$
• Coefficient of $x$: $-A + 2B - C - 2D = 0$
• Constant term: $-A + B + C + D = 1$

We already found $B = 1/2$ and $D = 1/2$. Substituting these values into the system of equations and solving, we find:

From $A + C = 0$, we have $C = -A$.

Substituting into $-A + B + C + D = 1$, we get $-A + 1/2 - A + 1/2 = 1$, which simplifies to $-2A = 0$, so $A = 0$.

Since $C = -A$, then $C = 0$.

Thus, we have found $A = 0$, $B = 1/2$, $C = 0$, and $D = 1/2$.

Reconstructing the Partial Fractions

With the constants A, B, C, and D now determined, we can reconstruct the partial fraction decomposition. Recall the form we set up earlier:

$(x^2 + 1)/((x^2 - 1)^2) = A/(x - 1) + B/(x - 1)^2 + C/(x + 1) + D/(x + 1)^2$

Substituting the values we found, $A = 0$, $B = 1/2$, $C = 0$, and $D = 1/2$, we get:

$(x^2 + 1)/((x^2 - 1)^2) = 0/(x - 1) + (1/2)/(x - 1)^2 + 0/(x + 1) + (1/2)/(x + 1)^2$

Simplifying, we obtain the final partial fraction decomposition:

$(x^2 + 1)/((x^2 - 1)^2) = 1/(2(x - 1)^2) + 1/(2(x + 1)^2)$

This result expresses the original rational function as a sum of simpler fractions, each with a squared linear factor in the denominator. This decomposition is invaluable for various applications, such as integration in calculus. The process of reconstructing the partial fractions from the solved constants demonstrates the power and utility of this technique. By breaking down a complex rational function into simpler components, we make it more amenable to analysis and manipulation. This final step solidifies our understanding of partial fraction decomposition and its role in simplifying algebraic expressions.

Applications and Significance

The utility of partial fraction decomposition extends far beyond mere algebraic manipulation. It is a cornerstone technique in various fields, notably calculus and engineering. In calculus, partial fraction decomposition is indispensable for integrating rational functions. When faced with an integral of a complex rational function, decomposing it into partial fractions allows us to express the integral as a sum of simpler integrals, which are often straightforward to evaluate. For example, the decomposition we derived, $(x^2 + 1)/((x^2 - 1)^2) = 1/(2(x - 1)^2) + 1/(2(x + 1)^2)$, transforms a seemingly complex integral into two manageable integrals. This simplification is crucial for solving many calculus problems, particularly those involving rational functions.

In engineering, partial fraction decomposition plays a vital role in system analysis, particularly in electrical engineering and control systems. Transfer functions, which describe the input-output relationship of a system, are often rational functions. Decomposing these transfer functions into partial fractions facilitates the analysis of system stability, response, and behavior. Each partial fraction corresponds to a mode of the system, allowing engineers to understand and control the system's dynamics. Moreover, in signal processing, partial fraction decomposition is used to analyze and synthesize signals by breaking them down into simpler components. The ability to simplify complex rational functions into manageable parts is a powerful tool for engineers, enabling them to design and analyze systems effectively.

The significance of partial fraction decomposition also lies in its ability to simplify complex algebraic expressions, making them easier to work with. This simplification is not only beneficial for manual calculations but also for numerical computations and computer algebra systems. By reducing the complexity of rational functions, partial fraction decomposition enhances the efficiency and accuracy of mathematical and engineering analyses. Its wide range of applications underscores its importance as a fundamental technique in mathematics and its related disciplines.

Conclusion

In summary, the process of partial fraction decomposition is a powerful and versatile technique for simplifying rational functions. Through this detailed exploration of decomposing $(x^2 + 1)/((x^2 - 1)^2)$, we have seen how to systematically break down a complex rational function into a sum of simpler fractions. The key steps involve factoring the denominator, setting up the appropriate form of the partial fractions, clearing denominators, solving for the unknown constants, and reconstructing the decomposed expression. The substitution method and the method of equating coefficients provide effective means to solve for these constants, each offering a unique approach to the problem.

The significance of partial fraction decomposition extends beyond algebraic manipulation, finding critical applications in calculus, engineering, and other scientific disciplines. In calculus, it simplifies the integration of rational functions, while in engineering, it aids in system analysis and design. The ability to transform complex expressions into simpler forms makes partial fraction decomposition an indispensable tool for mathematicians, engineers, and scientists alike.

By mastering this technique, one gains a deeper understanding of rational functions and their properties. The systematic approach to decomposition fosters analytical thinking and problem-solving skills, essential for success in various fields. As we have demonstrated, partial fraction decomposition is not just a mathematical procedure; it is a gateway to simplifying complexity and unlocking solutions in a wide range of applications. The elegance and efficiency of this technique underscore its importance in the mathematical toolkit, making it a valuable asset for anyone engaged in mathematical analysis and problem-solving.