Integrating Rational functions by the Method of Partial fraction Decomposition. Antony L. Foster

Integrating Rational functions by the Method of Partial fraction Decomposition By Antony L. Foster At times, especially in calculus, it is necessary, it is necessary to express a fraction as the sum of two or more others that are simpler in form than the original. The simpler fractions thus obtained are called partial fractions. In this discussion, we shall consider the problem of expressing a given fraction as the sum of partial fractions. A rational function rr(xx) = NN(xx)/DD(xx) is the quotient of two polynomials NN(xx) aaaaaa DD(xx). We shall deal exclusively with rational functions in this discussion, and we shall develop methods that apply only to proper rational fractions, that is, those rational functions in which the numerator is of lower degree than the denominator. In algebra, it is known that we can express any polynomial as the product of integral powers of linear and irreducible quadratic factors. Consequently, every rational fraction belongs to one of the following four cases: Case 1: All factors of the denominator are linear and none of them are repeated. Case 2: All factors of the denominator are linear and some of them are repeated. Case 3: The denominator contains irreducible quadratic factors none of which are repeated. Case 4: The denominator contains irreducible quadratic factors some of which are repeated. We shall employ the following theorem in the next four paragraphs, each of which concerns one of the four cases. The proof of the theorem is omitted because it is beyond the scope of this discussions. THEOREM: If a proper rational function rr(xx) = NN(xx) in lowest terms is expressed as the sum of partial fractions, then: DD(xx) 1. To every linear factor aaaa + bb of the denominator DD(xx) that appears without repetition, there corresponds a partial fractions AA AA aaaa + bb = aa xx + bb aa where AA is a constant. 2. To every factor (aaaa + bb) kk of the denominator DD(xx), there correspond the partial fractions AA 1 aaaa + bb, AA 2 (aaaa + bb) 2, AA 3 (aaaa + bb) 3,, AA kk (aaaa + bb) kk

where AA 1, AA 2,, AA kk are constants. 3. To every irreducible quadratic factor aaxx 2 + bbbb + cc of the denominator DD(xx)that appears without repetition, there corresponds the partial fraction where AA aaaaaa BB are constants. AAAA + BB aaxx 2 + bbbb + cc 4. If aaxx 2 + bbbb + cc is irreducible, then to every factor (aaxx 2 + bbbb + cc) kk of the denominator DD(xx), there correspond the partial fractions AA 1 xx + BB 1 aaxx 2 + bbbb + cc, AA 2 xx + BB 2 (aaxx 2 + bbbb + cc) 2,, AA kk xx + BB kk (aaxx 2 + bbbb + cc) kk where AA 1, AA 2,, AA kk, BB 1, BB 2,, BB kk are constants. CASE 1 The Denominator DD(xx) of the proper rational function contains only distinct linear factors: Example: Express the function rr(xx) = NN(xx) DD(xx) = (2xx 2 +xx+1) (xx+2)(3xx+1)(xx+3) (1) as the sum of partial fractions. Solution: There are two methods of solution, the first of them is as follows: METHOD I: Each factor of the denominator of rr(xx) is linear and appears only once. Hence, by part 1 of the theorem above, the partial fractions are AA xx + 2, BB 3xx + 1, CC xx + 3 Thus, we have rr(xx) = 2xx 2 +xx+1 (xx+2)(3xx+1)(xx+3) = AA + BB + CC xx+2 3xx+1 xx+3 (2) in which AA, BB, aaaaaa CC are to be determined so that (2) holds for every value of xx except possibly for those that cause one of the denominators to vanish. If we multiply each member of (2) by the LCD (xx + 2)(3xx + 1)(xx + 3), we get 2xx 2 + xx + 1 = AA(3xx + 1)(xx + 3) + BB(xx + 2)(xx + 3) + CC(xx + 2)(3xx + 1) (3)

By performing the indicated multiplications and collecting the terms in (3), we have 2xx 2 + xx + 1 = (3AA + BB + 3CC)xx 2 + (10AA + 5BB + 7CC)xx + (3AA + 6BB + 2CC) (4) If AA, BB, aaaaaa CC are determined so that 3AA + BB + 3CC = 2 10AA + 5BB + 7CC = 1 3AA + 6BB + 2CC = 1 then both sides of (3) are identical, and, therefore, the equation is satisfied for every value of xx. Consequently, for the values of AA, BB, aaaaaa CC thus determined, (2) is true for every value of xx, with the possible exception of those for which a denominator vanishes. We can solve this system of three linear equations in AA, BB, aaaaaa CC by methods learned in your elementary algebra courses (make sure you know how to solve linear systems like that above) and obtain that AA = 7, BB = 1, CC = 2. 5 5 Therefore, 2xx 2 + xx + 1 (xx + 2)(3xx + 1)(xx + 3) = 7 5 (xx + 2) + 1 5 (3xx + 1) + 2 xx + 3 (5) METHOD II: We shall present a second method for determining AA, BB, aaaaaa CC so that both sides of (2) are equal for all values of xx except possibly for xx = 2, xx = 3, xx = 1 for which one of the denominators vanishes. 3 If the members of (2) are equal for all values of xx with the possible exception of the three mentioned above, then the members of (4) are equal for all values of xx including these three. Since the right members of (3) and (4) are two forms of the same polynomial, the two members of (3) are equal for all values of xx. If we let xx = 2, the coefficients of BB aaaaaa CC in (3) vanish, and we have 8 2 + 1 = ( 5)(1)AA 5AA = 7 AA = 7 5 Similarly, when xx = 1, (3) becomes 3 2 9 1 3 + 1 = 5 3 8 BB 3 2 3 + 9 = 40BB 40BB = 8 BB = 1 5 Finally, if we let xx = 3, we have 18 3 + 1 = 8CC 8CC = 16 CC = 2

Hence, we have (5). The use of the second method in case 1 enables us to avoid solving the system of equations which involves AA, BB, aaaaaa CC as unknowns above and thus usually saves time and labor. It can frequently be employed to an advantage in other cases; in particular, if a linear factor appears in the denominator of rr(xx) FORMULA 1 AA αααα + ββ = AA ln αααα + bb + CC αα Illustration 1: Now in calculus, if we were asked to evaluated the indefinite integral 2xx 2 + xx + 1 (xx + 2)(3xx + 1)(xx + 3) Then expressing the integrand as a sum of partial fractions as shown in (5) above, we can then write 2xx 2 + xx + 1 (xx + 2)(3xx + 1)(xx + 3) 7 = 5 (xx + 2) + 1 5 (3xx + 1) + 2 xx + 3 = 7 5 1 xx + 2 + 1 15 3 1 + 2 3xx + 1 xx + 3 = 7 1 ln xx + 2 + ln 3xx + 1 + 2 ln xx + 3 + CC 5 15

EXERCISES FOR DISTINCT LINEAR FACTORS (CASE 1) Evaluate the indefinite integrals below by decomposing each of the integrands as a sum of partial fractions. These will be Collected! 1. 5xx+4 (xx 1)(xx+2) 2. 2xx+5 (xx+1)(xx+3) 3. 8tt 7 (2tt 1)(tt 2) 4. 5yy (yy+1)(2yy 3) 5. ss 8 (3ss+2)(2ss 3) 6. 8xx 31 (xx+3)(2xx 5) 7. 2ww+7 (2ww+5)(ww+3) 8. 5yy 3 (yy 5)(2yy+1) 9. 16qq+12 4qq 2 +4qq 3 10. xx+8 15xx 2 7xx 2

11. 19xx 46 6xx 2 29xx+35 12. 23xx 10 15xx 2 +xx 6 13. 2xx 2 +4xx 6 2xx 2 3xx 2 14. 6xx 2 +9xx 1 3xx 2 +2xx 1 15. 6xx 2 +3xx 11 2xx 2 +xx 6 16. 6xx 2 14xx+5 6xx 2 13xx+6 17. 5yy 2 +5yy 4 (yy 1)(yy+1)(yy+2) 18. 4ss2 +2ss 18 (ss+2)(ss 1)(ss+3) 19. 3tt 2 13tt 28 (tt+1)(tt+2)(tt 3) 20. 4θθ 2 +7θθ 6 (θθ 2)(θθ+2)(θθ+1) 21. 12xx 2 7 (XX 1)(2xx 1)(2xx+3) 22. 17ww 2 +18ww 72 (ww+1)(2ww+5)(3ww 1) 23. 6ss 2 +25ss+6 (ss 3)(2ss 1)(2ss+3) 24. 7YY 2 47YY 72 (YY+5)(3YY+2)(2YY 3)

CASE 2 The Denominator DD(xx) of the proper rational function contains only linear factors some of which repeat. If the denominator of a proper rational function in factored form contains only linear factors, but one or more are repeated, we use the method illustrated in the example below for expressing it as the sum of partial fractions. Example: Decompose into partial fractions. RR(xx) = 3xx 2 + 5xx + 1 (xx 1)(xx + 2) 2 Solution: Again we have available two methods of solution; the following is the first one. By Part 1 of the theorem, we must have the partial fraction AA xx 1 corresponding to the linear factor xx 1 of the denominator DD(xx) = (xx 1)(xx + 2) 2, and by part 2 of the theorem, we must also have the partial fractions BB xx + 2, CC (xx + 2) 2 corresponding to the factor (xx + 2) 2. Hence we have 3xx 2 + 5xx + 1 (xx 1)(xx + 2) 2 = AA xx 1 + BB xx + 2 + CC (xx + 2) 2 (1) and we must find the values of AA, BB, aaaaaa CCso that the member of (1) are equal for all values of xx, with the possible exception of xx = 1 aaaaaa xx = 2. The first step in this process is to multiply each member of (1) by (xx 1)(xx + 2) 2 and get 3xx 2 + 5xx + 1 = AA(xx + 2) 2 + BB(xx 1)(xx + 2) + CC(xx 1) (2) 3xx 2 + 5xx + 1 = (AA + BB)xx 2 + (4AA + BB + CC)xx + 4AA 2BB CC (3) The members of (3) will be equal for all values of xx if AA, BB, aaaaaa CC are determined so that the coefficients of the like powers of xx in the right and left members are equal. By equating the coefficients of xx 2, xx, and the two constant terms, we obtain the following system of three linear equations in AA, BB, aaaaaa CC: AA + BB = 3 4AA + BB + CC = 5 4AA 2BB CC = 1 The solutions of this system of equations is AA = 1, BB = 2, aaaaaa CC = 1. Therefore,

3xx 2 + 5xx + 1 (xx 1)(xx + 2) 2 = 1 xx 1 + 2 xx + 2 1 (xx + 2) 2 The second method of evaluating AA, BB, aaaaaa CC is similar to the second method applicable to case 1. If the members of (2) are equal for all values of xx, with the possible exception of xx = 1 aaaaaa xx = 2, then the members of (2) are equal for all values of xx. If we let xx = 1, in (2), the coefficients of BB aaaaaa CC vanish, and we have 3 + 5 + 1 = 3 2 AA 9AA = 9 AA = 1 If xx = 2, the coefficients of AA aaaaaa BB vanish in (2), and we have 12 10 + 1 = 3CC 3CC = 3 CC = 1 Since there is no single value of xx for which the coefficients of AA aaaaaa CC vanish simultaneously in (2), we must resort to another method to evaluate BB. Since the members of (2) hold for all values of xx, we may substitute any convenient value for this variable together with the values of AA aaaaaa CC above in (2) and obtain an equation containing only BB. So if we let xx = 0, AA = 1, aaaaaa CC = 1, (2) becomes 3(0) 2 + 5(0) + 1 = (1 4) + ( 1)(2)BB + ( 1)( 1) 1 = 5 2BB 2BB = 4 BB = 2 FORMULA 2 AA AA = + CC, kk 2 (αααα + ββ) kk αα(1 kk)(αααα + ββ) kk 1 Illustration 2: Thus, in calculus, if we were asked to evaluate the indefinite integral 3xx2 + 5xx + 1 (xx 1)(xx + 2) 2 Then expressing the integrand as a sum of partial fractions would give us 3xx2 + 5xx + 1 1 = (xx 1)(xx + 2) 2 xx 1 + 2 xx + 2 1 (xx + 2) 2 = 1 1 + 2 xx 1 xx + 2 1 (xx + 2) 2 = ln xx 1 + 2 ln xx + 2 + 1 xx + 2 + CC

EXERCISES FOR REPEATED LINEAR FACTORS (CASE 2) Evaluate the indefinite integrals below by decomposing each of the integrands as a sum of partial fractions. These will be Collected! 1. 2tt+5 (tt+1) 2 2. 2xx+11 (xx 4) 2 3. 3xx+1 (3xx 1) 2 4. 4xx 5 (2xx 1) 2 5. 27xx 2 +21xx+8 (3xx+2) 3 6. 4xx 2 5xx+1 (xx+1) 3 7. 16zz 2 +54zz 40 (2zz 3) 3 8. 20ss2 144ss+265 (2ss 7) 3 9. 12ww 2 9ww+20 (2ww+3)(3ww 1) 2 10. 8xx 2 +35xx+9 (2xx 1)(xx 4) 2 11. 24xx 2 94xx+88 (2xx 3)(3xx 5) 2 12. 22TT2 +32TT+138 (5TT+2)(2TT 7) 2 13. 3yy 3 9yy 2 +49yy 25 (2yy 1)(yy+2)(yy 1) 2

14. 3xx 3 11xx 2 18xx +46 (xx+3)(xx 2)(xx 1) 2 15. 14ss3 +14ss 2 +ss 7 (2ss+1)(ss 4)(ss+1) 2 16. 2yy 3 5yy 2 27yy 24 (yy+1)(yy 1)(yy+2) 2 17. 2xx 2 +5xx+1 (xx+1) 2 18. 3tt 2 16tt+20 (tt 3) 2 19. 2zz 3 10zz 2 +4zz+11 (2zz+1)(zz 2) 2 20. 3xx 3 12xx 2 3xx+10 (3xx 2)(xx 2) 2

CASE 3 The Denominator DD(xx) of the proper rational function contains distinct irreducible quadratic factors. If the first power of an irreducible quadratic function appears among the factors of the denominator of a rational function that is to be decomposed into partial fractions, it must appear as the denominator of one of the partial fractions. The numerator of the partial fraction that has the quadratic denominator must be a linear function. The linear factors of the denominator enter exactly as in the previous two cases (case 1 and case 2). Example 1: Decompose FF(xx) = 14xx 3 + 14xx 2 4xx + 3 (3xx 2 xx + 1)(xx 1)(xx + 2) into partial fractions. Solution: The quadratic factor 3xx 2 xx + 1 is irreducible in the real numbers; hence it must be used as the denominator of a partial fraction that has a linear function AAAA + BB for its numerator. Each of the linear factors will enter as a denominator of a partial fraction with a constant numerator. Thus, 14xx 3 + 14xx 2 4xx + 3 (3xx 2 xx + 1)(xx 1)(xx + 2) = AAAA + BB 3xx 2 xx + 1 + CC xx 1 + DD xx + 2 (1) In order to find the values of AA, BB, CC, aaaaaa DD so that the members of (1) are equal, we first multiply each member of (1) by (3xx 2 xx + 1)(xx 1)(xx + 2) and get 14xx 3 + 14xx 2 4xx + 3 = (AAAA + BB)(xx 1)(xx + 2) + CC(3xx 2 xx + 1)(xx + 2) + DD(3xx 2 xx + 1)(xx 1) (2) 14xx 3 + 14xx 2 4xx + 3 = (AA + 3CC + 3DD)xx 3 + (AA + BB + 5CC 4DD)xx 2 + ( 2AA + BB CC + 2DD)xx + ( 2BB + 2CC DD) (3) We may now obtain, by equating the coefficients of equal powers of xx, the four linear equations in AA, BB, CC, and DD which follow: AA + 3CC + 3DD = 14 AA + BB + 5CC 4DD = 14 2AA + BB CC + 2DD = 4 2BB + 2CC DD = 3 The solution of this system can be obtained by Gaussian Elimination (which can be found in Precalculus and linear algebra textbooks) or by solving the last equation for DD in terms of BB and CC, replacing each DD in the other three equations by this value, solving the resulting system of three linear equations in three unknowns, and finally, obtaining the value of DD from the last equation. The solution thus obtained is AA = 2, BB = 1, CC = 3, DD = 1. Hence,

14xx 3 + 14xx 2 4xx + 3 FF(xx) = (3xx 2 xx + 1)(xx 1)(xx + 2) = 2xx + 1 3xx 2 xx + 1 + 3 xx 1 + 1 xx + 2 Example 2: Decompose rr(xx) = 4xx 4 + 4xx 3 xx 2 + xx + 1 (xx 2 + xx + 1)(xx 2 xx 3)(xx + 1) into partial fractions. Solution 4xx 4 + 4xx 3 xx 2 + xx + 1 (xx 2 + xx + 1)(xx 2 xx 3)(xx + 1) = AAAA + BB xx 2 + xx + 1 + CCCC + DD xx 2 xx 3 + EE xx + 1 (1) After multiplying each member of (1) by (xx 2 + xx + 1)(xx 2 xx 3)(xx + 1), we have 4xx 4 + 4xx 3 xx 2 + xx + 1 = (AAAA + BB)(xx 2 xx 3)(xx + 1) + (CCCC + DD)(xx 2 + xx + 1)(xx + 1) + EE(xx 2 + xx + 1)(xx 2 xx 3) (2) = (AA + CC + EE)xx 4 + (BB + 2CC + DD)xx 3 + ( 4AA + 2CC + 2DD 3EE)xx 2 + ( 3AA 4BB + CC + 2DD 4EE)xx + ( 3BB + DD 3EE) (3) We can obtain the following system of equations by equating the coefficients of the equal powers of xx: AA + CC + EE = 4 BB + 2CC + DD = 4 4AA + 2CC + 2DD 3EE = 1 3AA 4BB + CC + 2DD 4EE = 1 3BB + DD 3EE = 1 The solution of this system can be obtained by Gaussian elimination or by obtaining an expression for DD in terms of BB aaaaaa EE from the last equation, substituting this for DD in the other four equations, thus getting a system of four linear equations in four unknowns, and solving the system by any of the available methods. The method suggested in Example 1 can be used. The solution is AA = 1, BB = 1, CC = 2, DD = 1, EE = 1. Therefore, 4xx 4 + 4xx 3 xx 2 + xx + 1 (xx 2 + xx + 1)(xx 2 xx 3)(xx + 1) = xx 1 xx 2 + xx + 1 + 2xx + 1 xx 2 xx 3 + 1 xx + 1

FORMULA 3 2 2αααα + ββ 4αααα ββ 2 tan 1 4αααα ββ 2 + CC iiii ββ2 4αααα < 0 1 ααxx 2 + ββββ + γγ = 1 ββ 2 4αααα ln 2αααα + ββ ββ2 4αααα 2αααα + ββ + ββ 2 4αααα + CC iiii ββ2 4αααα > 0 Illustration 3: Find the indefinite integral FORMULA 4 xx ααxx 2 + ββββ + γγ = 1 2αα ln ααxx2 + ββββ + γγ ββ 2αα 1 ααxx 2 + ββββ + γγ 14xx3 + 14xx 2 4xx + 3 (3xx 2 xx + 1)(xx 1)(xx + 2) Solution: By expressing the integrand as a sum of partial fractions as seen in Example 1 just above, we have 14xx3 + 14xx 2 4xx + 3 (3xx 2 xx + 1)(xx 1)(xx + 2) = 2xx + 1 3xx 2 xx + 1 + 3 xx 1 + 1 xx + 2 = 2xx + 1 1 1 3xx 2 + 3 + xx + 1 xx 1 xx + 2 = 2xx + 1 3xx 2 + 3 ln xx 1 + ln xx + 2 + CC xx + 1 xx = 2 3xx 2 xx + 1 + 1 3xx 2 xx + 1 + 3 ln xx 1 + ln xx + 2 + CC = 1 3 ln 3xx2 xx + 1 + 4 3 1 3xx 2 + 3 ln xx 1 + ln xx + 2 + CC xx + 1 = 1 3 ln 3xx2 xx + 1 + 8 3 11 tan 1 6xx 1 + 3 ln xx 1 + ln xx + 2 + CC 11

Illustration 4 : Suppose that in calculus we are asked to find 4xx 4 + 4xx 3 xx 2 + xx + 1 (xx 2 + xx + 1)(xx 2 xx 3)(xx + 1) Solution: By decomposing the integrand as a sum of partial fractions (using case 1, case 3) as shown in Example 2 above, we have 4xx 4 + 4xx 3 xx 2 + xx + 1 (xx 2 + xx + 1)(xx 2 xx 3)(xx + 1) = xx 1 xx 2 + xx + 1 + 2xx + 1 xx 2 xx 3 + 1 xx + 1 = xx 1 2xx + 1 1 xx 2 + + xx + 1 xx 2 + xx 3 xx + 1 = 1 2 ln xx2 + xx + 1 3 tan 1 2xx + 1 3 + 1 ln xx2 xx 3 + 2 xx 2 xx 3 + ln xx + 1 + CC = 1 2 ln xx2 + xx + 1 3 tan 1 2xx + 1 3 + ln xx2 xx 3 + 2 13 1 13 ln 2xx + ln xx + 1 + CC 2xx 1 + 13

EXERCISES FOR DISTINCT IRREDUCIBLE QUADRATIC FACTORS (CASE 3) Find the following indefinite integral by decomposing the integrand into partial fractions 1. 3xx 2 10xx+16 (xx 3)(xx 2 +xx+1) 2. 2xx+5 (xx 1)(xx 2 +2) 3. 5xx 2 +xx+2 (xx+1)(xx 2 +1) 4. xx 2 4xx 3 (2xx 3)(xx 2 xx 3) 5. 5xx 3 +14xx 2 +71xx+14 (xx+5)(2xx 1)(xx 2 3) 6. xx 3 +11xx 2 +13xx 5 (xx 3)(3xx+1)(xx 2 xx+2) 7. 9xx 3 +19xx 2 +2xx+3 (2xx 3)(xx+2)(xx 2 +3) 8. 5xx 3 31xx 2 3xx 23 (3xx 1)(xx 3)(xx 2 +3xx+4) 9. 3xx 3 10xx 2 +9xx 6 (xx 1) 2 (xx 2 +1) 10. xx 3 +8xx 2 +2xx 19 (xx 2) 2 (xx 2 +5) 11. 4xx 3 11xx 2 +12xx 12 (xx 2) 2 (xx 2 +xx+1) 12. 10xx 3 +3xx 2 19xx +12 (2xx 1) 2 (xx 2 +xx 3) 13. 2xx 3 8xx+4 (xx 2 +2)(xx 2 2)

14. 3xx 3 3xx 2 4xx 14 (xx 2 5)(xx 2 xx+3) 15. xx 3 13xx 2 +9 (xx 2 +3)(xx 2 3xx 3) 16. 3xx 3 10xx 2 +7xx 3 (xx 2 +1)(xx 2 3xx 1) 17. 6xx 3 9xx 2 +33xx 64 (2xx 3)(xx 2 +5) 18. 2xx 3 10xx 2 +11xx +19 (xx 5)(2xx 2 3xx+2) 19. xx 3 +4xx 2 7xx+6 (xx 1)(xx 2 +1) 20. 2xx 3 +4xx 2 11xx 19 (xx+4)(xx 2 3)

CASE 4 The Denominator DD(xx) of the proper rational function contains repeated irreducible quadratic factors. The final case to be considered (case 4) is that in which the factors of the denominator of the given proper rational function contains powers of one or more irreducible quadratic factors. By part 4 of the theorem in the beginning of this discussion, to every factor of the type (ααxx 2 + ββββ + γγ) kk appearing in the denominator of a rational function there corresponds the partial fractions where AA 1, AA 2,, AA kk, BB 1, BB 2,, BB kk are constants. AA 1 xx + BB 1 aaxx 2 + bbbb + cc, AA 2 xx + BB 2 (aaxx 2 + bbbb + cc) 2,, AA kk xx + BB kk (aaxx 2 + bbbb + cc) kk The linear and nonrepeated quadratic factors of the given denominator enter in the same manner as in cases 1 to 3. The numerator of each partial fraction whose denominator contains a quadratic factor should be a linear function. Example: Decompose the following fraction into partial fractions. 6xx 4 + 11xx 3 + 18xx 2 + 14xx + 6 (xx + 1)(xx 2 + xx + 1) 2 Solution: The denominator contains a linear function and the square of an irreducible quadratic function as factors; hence we let 6xx 4 + 11xx 3 + 18xx 2 + 14xx + 6 (xx + 1)(xx 2 + xx + 1) 2 = AA xx + 1 + BBBB + CC xx 2 + xx + 1 + DDDD + EE (xx 2 + xx + 1) 2 and evaluate the undetermined constants after multiplying each member by (xx + 1)(xx 2 + xx + 1) 2. Thus 6xx 4 + 11xx 3 + 18xx 2 + 14xx + 6 = AA(xx 2 + xx + 1) 2 + (BBBB + CC)(xx + 1)(xx 2 + xx + 1) + (DDDD + EE)(xx + 1) = (AA + BB)xx 4 + (2AA + 2BB + CC)xx 3 + (3AA + 2BB + 2CC + DD)xx 2 + (2AA + BB + 2CC + DD + EE)xx + (AA + CC + EE). Therefore, by equating the coefficients of equal powers of xx, we obtain AA + BB = 6 2AA + 2BB + CC = 11 3AA + 2BB + 2CC + DD = 18 2AA + BB + 2CC + DD + EE = 14 AA + CC + EE = 6

We can solve this system by means of the method suggested in the solution of Example 2 in Case 3. The solution to this system of equations is AA = 5, BB = 1, CC = 1, DD = 3, aaaaaa EE = 2. Therefore, 6xx 4 + 11xx 3 + 18xx 2 + 14xx + 6 (xx + 1)(xx 2 + xx + 1) 2 = 5 xx + 1 + xx 1 xx 2 + xx + 1 + 3xx + 2 (xx 2 + xx + 1) 2 FORMULA 5 1 (ααxx 2 + ββββ + γγ) 2 = 2αααα + ββ (4αααα ββ 2 )(ααxx 2 + ββββ + γγ) + 2αα 4αααα ββ 2 1 ααxx 2 + ββββ + γγ FORMULA 6 xx (ααxx 2 + ββββ + γγ) 2 = ββββ + 2γγ (4αααα ββ 2 )(ααxx 2 + ββββ + γγ) ββ 4αααα ββ 2 1 ααxx 2 + ββββ + γγ 1 (ααxx 2 + ββββ + γγ) 3 = FORMULA 7 2αααα + ββ 2(4ααγγ ββ 2 )(ααxx 2 + ββββ + γγ) 2 + 6αα 2 xx + 3αααα (4αααα ββ 2 ) 2 (ααxx 2 + ββββ + γγ) + 6αα 2 (4ααγγ ββ 2 ) 2 1 ααxx 2 + ββxx + γγ FORMULA 8 xx (ααxx 2 + ββxx + γγ) 3 = ββxx + 2γγ 2(4ααγγ ββ 2 )(ααxx 2 + ββxx + γγ) 2 6ααββxx + 3ββ 2 2(4ααγγ ββ 2 ) 2 (ααxx 2 + ββxx + γγ) 3ααββ (4ααγγ ββ 2 ) 2 ααxx 2 + ββxx + γγ Illustration 5: Find the indefinite integral Solution 6xx4 + 11xx 3 + 18xx 2 + 14xx + 6 (xx + 1)(xx 2 + xx + 1) 2 6xx4 + 11xx 3 + 18xx 2 + 14xx + 6 (xx + 1)(xx 2 + xx + 1) 2 = 5 xx + 1 + xx 1 xx 2 + xx + 1 + 3xx + 2 (xx 2 + xx + 1) 2 = 5 1 xx 1 + xx + 1 xx 2 + xx + 1 + 3xx + 2 (xx 2 + xx + 1) 2 = 5 1 xx 1 + xx + 1 xx 2 + xx + 1 xx + 3 (xx 2 + xx + 1) 2 + 2 1 (xx 2 + xx + 1) 2 = 5 ln xx + 1 + 1 2 ln xx2 + xx + 1 3 tan 1 2xx + 1 3 + 3 2 2xx + 1 (xx 2 + xx + 1) 2 + 1 2 1 (xx 2 + xx + 1) 2

= 5 ln xx + 1 + 1 2 ln xx2 + xx + 1 3 tan 1 2xx + 1 3 3 2(xx 2 + xx + 1) + 1 2 1 (xx 2 + xx + 1) 2 = 5 ln xx + 1 + 1 2 ln xx2 + xx + 1 3 tan 1 2xx + 1 3 3 2(xx 2 + xx + 1) + 1 2 2xx + 1 3(xx 2 + xx + 1) + 4 3 3 tan 1 2xx + 1 3 + CC = 5 ln xx + 1 + 1 xx 4 2 ln xx2 + xx + 1 + 3(xx 2 + xx + 1) 7 3 3 tan 1 2xx + 1 + CC 3

EXERCISE FOR REPEATED QUADRATIC FACTORS (CASE 4) Find the value of each of the following integrals by decomposing the integrand in partial fractions. (These Problems will be Collected!) 1. 3xx 3 +4xx 5 (xx 2 +2) 2 2. xx 3 +2xx 2 3 (xx 2 2) 2 3. 2xx 3 +xx 2 +4xx+1 (xx 2 +xx+1) 2 4. 2xx 3 xx 2 6xx 8 (xx 2 xx 3) 2 5. xx 5 +2xx 4 3xx 2 3xx (xx 2 +xx 1) 3 6. xx 5 2xx 4 +4xx 3 2xx 2 (xx 2 xx+1) 3 7. xx 5 +2xx 3 xx+3 (xx 2 +1) 3 8. xx 3 +4xx 5 (xx 2 +2) 3 9. xx 4 xx 3 +1 xx 2 (xx 2 +1) 2 10. 2xx 4 +13xx 3 +13xx 2 12xx+2 xx 2 (xx 2 +3xx 1) 2 ddxx 11. xx 5 +xx 4 4xx+4 xx 2 (xx 2 xx+2) 2 12. 2xx 5 4xx 4 +12xx 3 +9xx 2 6xx+1 xx 2 (xx 2 +3xx 1) 2 13. 4xx 4 +xx 3 25xx 2 9xx+30 (xx+2)(xx 2 3) 2

14. 4xx 4 7xx 3 +5xx 2 xx+1 (2xx 1)(xx 2 xx+1) 2 15. 3xx 4 +2xx 3 +8xx 2 +7xx+2 (xx 1)(xx 2 +2) 2 16. xx 4 5xx 2 10xx 2 +26xx+33 (xx+3)(xx 2 xx 3) 2 THE END