Section 8.5 Reading Assignment: Integration of Rational Functions by Partial Fractions

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Section 8.5 Reading Assignment: Integration of Rational Functions by Partial Fractions Instructions. Read through this assignment and complete the three exercises below by reading the appropriate passages of the textbook. Partial fraction decomposition is an algebraic technique that may haye come up during precalculus. For us, it's useful for integrating rational functions. This technique is also used during differential equations as well. which may be worth keeping your work from this section if you plan on taking that class. We will be referring to the phrase \"undetermined coefficients\" in exercises, so it's worth keeping it its definition {defined near the bottom of p. 485]. Notice that these coefficients are unknown numbers in the partial fraction decomposition with any dependence on x written separately. Example 1 (p. 486 43?} is a good example of a base case. There are multiple ways of solying for the undetermined coefficients but obtaining a system of equations using the coefficients works for all situations, which makes it a consistent method. Each ofthe exercises here describes a special case that makes the process different from the basic case. It's worth keeping each ofthese special cases in mind when setting up these problems. Exercise 1. Read Example 2 [p. 48?]. Explain why the partial fraction takes the form it does. Describe the properties of the rational function that make this particular form necessary. Do not explain how to find the undetermined coefficients, just the general form ofthe partial fraction decomposition. Exercise 2. Read Example 3 [p. 488]. Explain what it means for a rational function to be \"improper\" and what must be done first in the case of an improper rational function before taking the partial fraction decomposition. Exercise 3. Read Example :1 ip. 483 489}. Explain why the partial fraction takes the form it does. Describe the properties of the rational function that make this particular form necessary. Do not explain how to find the undetermined coefficients, just the general form ofthe partial fraction decomposition. The last subsection \"Other 1ilv'ays to Determine the Coefficients\" (p. 490 491} describes alternative methods for finding the undetermined coefficients. Example 5 (p. 49D) is a long version of Heaviside's cover-up method.1 Example If (p. 490} and Example 3 {p. 491}, like Heaviside's method, uses the fact that you can plug in various values ofx to solve for the undetermined coefficients. Example 6 and this last subsection are optional. You can solve for the undetermined coefficients in any manner that you prefer. 1 For further information about this method, including more advanced examples, check out this [E from MIT. The class is differential equations, but this pdf itself does not require any differential equations to understand. Chapter 8 Techniques of Integration 8.5 Integration of Rational Functions by Partial Fractions 487 Solution Note that each of the factors (x - 1), (x + 1), and (x + 3) is raised only to the first power. Therefore, the partial fraction decomposition has the form x' + 4x + 1 B (x - D)(x + 1)(x + 3) * + 3 To find the values of the undetermined coefficients A. B, and C, we clear fractions and get x'+ 4 + 1 - A(x + 1)( + 3) + B(x - D)( + 3) + C(x - 1)(x + 1) = A(x' + 4r + 3) + 8(x' + 2x - 3) + c(x - 1) = (A + B + Cix + (4A + 28)x + (3A - 38 - q). The polynomials on both sides of the above equation are identical, so we equate coeffi cients of like powers of x, obtaining Coefficient of r-: A+ B+C=1 Coefficient of r': 4A + 2B 4 Coefficient of x": 3A - 38 - C = 1 There are several ways of solving such a system of linear equations for the unknowns A, B. and C, including elimination of variables or the use of a calculator or computer. The solu- tion is A = 3/4, 8 = 1/2, and C = -1/4. Hence we have x- + 4x + 1 dx = ?mlx - 1/ + ; In/x + 1| - -In/x+ 3|+ K. where & is the arbitrary constant of integration (we call it K here to avoid confusion with the undetermined coefficient we labeled as C).EXAMPLE 2 Use partial fractions to evaluate br + 7 (x + 2)] dr. Solution First we express the integrand as a sum of partial fractions with undetermined coefficients. fr + 7 B + *+2 Two terms because (x + 2) is squared (x + 2) (x + 2)2 6x + 7 - A(x + 2) + B Multiply both sides by ( + 2)2. = Ax + (2A + B) Equating coefficients of corresponding powers of x gives A = 6 and 2A + B = 12 + B = 7, or A = 6 and B = -5. Therefore. 6x + 7 (x + 23 x = 6 5 x+ 2 (x + 2)2/ d.x - 6/x42 - 5/(x +21" dx = 6In x + 2 + 5(x + 2)1 + C. The next example shows how to handle the case when f()/g(x) is an improper frac- tion. It is a case where the degree of f is larger than the degree of g.Chapter 8 Techniques of Integration 488 Chapter 8 Techniques of Integration EXAMPLE 3 Use partial fractions to evaluate 2 - 4x - x - 3dx. x - 2x - 3 Solution First we divide the denominator into the numerator to get a polynomial plus a proper fraction. 2x x - 2x - 3)2x3 - 4x - x - 3 2x3 - 4x' - 6x - 3 5x - 3 Then we write the improper fraction as a polynomial plus a proper fraction. 2x3 - 4x - x - 3 = 2r + 5x - 3 x - 2r - 3 x - 2r - 3 We found the partial fraction decomposition of the fraction on the right in the opening example, so (203 - 4x - x- 3dx = / 2xax + / 5x - 3 x - 2x - 3 x - 2x - 3 x - 2x dx +/ 3 I - 3d =x+ 2ln x+ 1|+ 3In|x - 3| + C.EXAMPLE 4 Use partial fractions to evaluate -2x + 4 ( + 1)6 - 1) dx. Solution The denominator has an irreducible quadratic factor x2 + 1 as well as a repeated linear factor (x - 1), so we write -2x + 4 Ax + B D ( + 1)(x - 1) (2) x+1 X (x - Clearing the equation of fractions gives -2r + 4 = (Ax + D)(x - 1) + ((x - D(x + 1) + D(x] + 1) = (A + C)x]+ (-24 + B - C+ Dix + (A - 28 + Cix + (B - C + D). Equating coefficients of like terms gives Coefficients of x*: O = A+C Coefficients of r: 0 = -24 +B - C+D Coefficients of x : -2 = A - 28 + C Coefficients of x"- 4 =B - C+D We solve these equations simultaneously to find the values of A, B. C, and D: -4 = -24. A = 2 Subtract fourth equation from second. C = -4 = -2 From the first equation B = (A + C+ 2)/2 =1 From the third equation and C = -A D =4- B+C=1. From the fourth equationChapter 8 Techniques of Integration 8.5 Integration of Rational Functions by Partial Fractions 489 We substitute these values into Equation (2). obtaining -2x + 4 2x + 1 _ _ 2 + (x + 1)(x - 1)3 x- 1 (x - 172 Finally, using the expansion above we can integrate: -2x + 4 2 (x] + 1)(x - 1)2 - 1) + ; = In (r + 1) + tan~] x - 2 In |x - 1/ - T+ C. X - 1 EXAMPLE 5 Use partial fractions to evaluate Solution The form of the partial fraction decomposition is 1 Bx + C , Dx + E x(x + 1) X "+1 (+1) Multiplying by x(x- + 1)', we have 1 = A(x + 1) + (Bx + Cu(x] + 1) + (Dx + Exx = A(x4 + 213 + 1) + B(x+ + x) + C(x' + x) + Dx' + Ex = (A + Bjx + Co] + (24 + B + Dj + (C+ Ex+ A. If we equate coefficients, we get the system A+B=0. C = 0. 2A +B+ D= 0, C+E=0. A=1.Solving this system gives A = 1, B = -1, C = 0, D = -1, and E = 0. Thus, dx + x(x + 1)2 r- + 1 (x] + 1)' dx x dx M=rtl, du = Zrer = In |x] - SIn |u| + + K = In |x] - - In (x2 + 1) + 1 + K HISTORICAL BIOGRAPHY 2(x3 + 1 ) Oliver Heaviside (1850-1925) = In - + K. www. goo . g1/ 5rnavz Vx+ 1 2(x+ 1) When the degree of the polynomial f(x) is less than the degree of g(x) and g(x) = (x - n)(x - m) . .(x - m) is a product of a distinct linear factors, each raised to the first power, there is a quick way to expand f(x)/g(x) by partial fractions.Chapter 8 Techniques of Integration 490 Chapter 8 Techniques of Integration EXAMPLE 6 Find A, B, and C in the partial fraction expansion x3 + 1 (x - 1)(x - 2)(x - 3) + B * - 2 (3) Solution If we multiply both sides of Equation (3) by (x - 1) to get B(x - 1) C(x - 1) (x - 2)(x - 3) - = A + x- 2 x- 3 and set & = 1, the resulting equation gives the value of A: (1) + 1 (1 - 2)(1 - 3) =A+0+0. A = 1. In exactly the same way, we can multiply both sides by (x - 2) and then substitute in x = 2. This gives (2) + 1 (2 - 1)(2 - 3) = B. So B = -5. Finally, we multiply both sides by (x - 3) and then substitute in * = 3. which yields (3) + 1 (3 - 1)(3 - 2) = C. and C = 5.Other Ways to Determine the Coefficients Another way to determine the constants that appear in partial fractions is to differentiate, as in the next example. Still another is to assign selected numerical values to .. EXAMPLE 7 Find A, B, and C in the equation B C (x + 1)3 x + 1 (x + 1)2 (x + 1)3 by clearing fractions, differentiating the result, and substituting x = -1. Solution We first clear fractions: x - 1 = A(x + 1) + B(x+ 1)+ c. Substituting x = -1 shows C = -2. We then differentiate both sides with respect to x, obtaining 1 = 2A(x + 1) + B. Substituting x = -1 shows B = 1. We differentiate again to get 0 = 24, which shows A = 0. Hence, (x + 1 (x + 1)- In some problems, assigning small values to x, such as x = 0, $1, 12, to get equa- tions in A, B, and C provides a fast alternative to other methods.Chapter 8 Techniques of Integration 8.5 Integration of Rational Functions by Partial Fractions 491 EXAMPLE 8 Find A, B, and C in the expression x3 + 1 B (x - 1)(x - 2)(x - 3) x - by assigning numerical values to x. Solution Clear fractions to get x' + 1 - A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2). Then let x = 1, 2, 3 successively to find A, B, and C. x=1: (1) + 1 = A(-1)(-2) + 8(0) + C(O) 2 = 24 A =1 x = 2: (2)- + 1 = A(0) + 8(1)(-1) + C(O) 5 - -B B =-5