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Chapter 6: Problem 48

More on Repeated Linear Factors The Heaviside Method is not very effective at finding the unknown numerators for par- tial fraction decompositions with repeated linear factors, but here is another way to find them. (a) If \(\frac{x^{2}+3 x+5}{(x-1)^{3}}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{(x-1)^{3}},\) show that \(A(x-1)^{2}+B(x-1)+C=x^{2}+3 x+5\) (b) Expand and equate coefficients of like terms to show that \(A=1,-2 A+B=3,\) and \(A-B+C=5 .\) Then find \(A, B\) , (c) Use partial fractions to evaluate \(\int \frac{x^{2}+3 x+5}{(x-1)^{3}} d x\)

Short Answer

Expert verified
The values for \(A\), \(B\), and \(C\) are \(1\), \(5\), and \(9\) respectively. The integral of \(\int \frac{x^{2}+3 x+5}{(x-1)^{3}} d x\), after substituting these values, evaluates to \(\ln|x-1| - \frac{5}{x - 1} - \frac{9}{2(x - 1)^{2}} + constant\).

Step by step solution

01

Expand the Left Hand Side of the Equation

Expand \(A(x-1)^{2}+B(x-1)+C\). First, expand \(A(x-1)^{2}\) to obtain \(Ax^{2} - 2Ax + A\). Similarly, expand \(B(x-1)\) to get \(Bx - B\). So, once simplified, the left side of the equation becomes \(Ax^{2} - 2Ax + A + Bx - B + C\).
02

Equate Coefficients of the Same Terms

From the left hand side of the equation we obtained in the previous step, now equate coefficients of like terms. This gives you three equations: \(A=1\) for the coefficients of \(x^{2}\), \(-2A + B = 3\) for the coefficients of \(x\) and \(A - B + C = 5\) for the constants.
03

Solve the System of Equations

We already have \(A = 1\) from equating the coefficients of \(x^{2}\). Substituting \(A = 1\) into \(-2A + B = 3\) gives \(B=3+2=5\). Substituting \(A = 1\) and \(B = 5\) into \(A - B + C = 5\) gives \(C=5-1+5=9\). So, \(A = 1\), \(B = 5\) and \(C = 9\).
04

Perform Partial Fraction Integration

Substitute the obtained values of \(A\), \(B\), and \(C\) into the integral \(\int \frac{x^{2}+3 x+5}{(x-1)^{3}} d x\) which simplifies to \(\int \left(\frac{1}{x - 1} + \frac{5}{(x - 1)^{2}} + \frac{9}{(x - 1)^{3}}\right)dx\). Now integrate term by term: \(\ln|x-1| - \frac{5}{x - 1} - \frac{9}{2(x - 1)^{2}} + constant\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heaviside Method
The Heaviside Method is a popular technique used in Partial Fraction Decomposition to find unknown numerators effectively. Named after the English mathematician Oliver Heaviside, this method is particularly useful when dealing with simple linear and non-repeated factors. However, it becomes less efficient with repeated linear factors, making it necessary to use alternative methods in such cases. The Heaviside Method primarily involves these steps:
  • Identifying the form of the partial fractions based on the factorization of the denominator.
  • Setting up equations by substituting strategic values that null out certain terms to solve for unknowns in simple cases.
But when the denominators are raised to powers greater than one, as in repeated linear factors, the method becomes cumbersome, necessitating a shift to more sophisticated algebraic techniques like equating coefficients.
Repeated Linear Factors
Repeated Linear Factors mean a single factor raised to a power within the denominator of a fraction. In Partial Fraction Decomposition, this scenario adds complexity, as we cannot simply apply the Heaviside Method. Instead, each term must be expanded into a series of partial fractions where each instance of the factor appears separately in descending power order. For example, given a denominator \( (x-1)^3 \), the decomposition will involve terms like:
  • \( \frac{A}{x-1} \)
  • \( \frac{B}{(x-1)^2} \)
  • \( \frac{C}{(x-1)^3} \)
Each coefficient must be determined using algebraic manipulation to equate the decomposed form back to the original function. This ensures that the expanded terms can correctly represent the initial function over all values of x.
Integration Techniques
Integration Techniques used in this exercise enable the integration of complex rational functions once they have been decomposed into simpler parts. By breaking down the original function into partial fractions, integration becomes straightforward.Here's how you integrate each common partial fraction term:
  • \( \int \frac{A}{x-1}dx \) integrates to \( A \ln|x-1| \).
  • \( \int \frac{B}{(x-1)^2}dx \) results in \( \frac{-B}{x-1} \).
  • \( \int \frac{C}{(x-1)^3}dx \) gives \( \frac{-C}{2(x-1)^2} \).
The integration process simplifies each fraction, highlighting the power and utility of breaking down polynomial fractions before integrating, thus easing the entire procedure.
Algebraic Manipulation
Algebraic Manipulation plays a crucial role in solving for the coefficients in Partial Fraction Decomposition, especially when dealing with repeated factors. It involves expanding, simplifying, and matching terms on both sides of an equation to find numerical values of the unknowns.The process typically includes:
  • Expanding expressions like \( A(x-1)^2 + B(x-1) + C \), using standard algebraic expansions to combine like terms.
  • Equating coefficients from both sides: Once expanded, each term's coefficient (like those for \( x^2, x, \) and constant terms) is set equal to the corresponding term in the original numerator.
  • Solving the resulting system of linear equations to discover the values of A, B, and C.
This systematic approach ensures accuracy and allows us to solve problems involving complex algebraic fractions with confidence.

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