Section 7.4
- 1. Chapter 7: Techniques of Integration Section 7.4: Integration of Rational Functions by Partial Fractions Alea Wittig SUNY Albany
- 2. Outline Introduction to Method of Partial Fractions Step 1: Polynomial Long Division Step 2: Factoring Q(x) Partial Fraction Decomposition Case I: Distinct Linear Factors Case II: Repeated Linear Factors Case III: Distinct Irreducible Quadratic Factors Case IV: Repeated Irreducible Quadratic Factors Rationalizing Substitutions
- 3. Introduction to Method of Partial Fractions
- 4. Rational Functions ➤ A rational function is a ratio of polynomials. f (x) = P(x) Q(x) = anxn + an−1xn−1 + . . . + a1x + a0 bmxm + bm−1xm−1 + . . . + b1x + b0 ➤ In this section we will learn that any given rational function can be broken down into a set of standard integrals, each of which can be computed routinely. ➤ As a warmup exercise, compute the following standard integrals: Z A x + a dx (a) Z B (x + a)2 dx (b)
- 5. Exercise a To compute Z A x + a dx (a) we first make the u sub: u = x + a, du = dx Z A x + a dx = Z A u du = A Z 1 u du = A ln |u| + C = A ln |x + a| + C
- 6. Exercise b To compute Z B (x + a)2 dx (b) we first make the u sub: u = x + a, du = dx Z B (x + a)2 dx = Z B u2 du = B Z 1 u2 du = B Z u−2 du = B u−2+1 −2 + 1 + C = B u−1 −1 + C = − B u + C = − B x + a + C
- 7. The Method of Partial Fractions The Method of Partial Fractions can be broken down into 3 steps: Step 1: Polynomial Long Division Step 2: Factorization of Q(x) Step 3: Partial Fraction Decomposition
- 8. Step 1: Polynomial Long Division
- 9. Proper vs. Improper f (x) = P(x) Q(x) ➤ f is called proper if the degree (highest power) of P is less than the degree of Q. eg, f (x) = x2 − 2x + 1 x3 + x ➤ f is called improper if the degree of P is greater than or equal to the degree of Q. eg, f (x) = 3x5 + x x2 − 2x + 1
- 10. Polynomial Long Division ➤ Any rational function f can be written in the form f (x) = P(x) Q(x) = S(x) |{z} polynomial + R(x) Q(x) | {z } proper rational ➤ If f is proper, then S(x) = 0 and R(x) = P(x) ➤ If f is improper then we must divide Q into P using long division until a remainder R(x) is obtained with deg(R) < deg(Q). ➤ Warning: Partial fraction decomposition (step 3) only works on proper rational functions.
- 11. Example 1 Video Example 1
- 12. Exercise c As an exercise, use long division to show that x3 + x x − 1 = x2 + x + 2 + 2 x − 1 Use this to compute the integral Z x3 + x x − 1 dx
- 13. ➤ In example 1 and exercise c we were able to write the rational function as a sum of standard integrals which were simple to compute. ➤ If we are not able to integrate after step 1 then move on to step 2.
- 14. Step 2: Factoring Q(x)
- 15. Factor Q(x) ➤ After Step 1 we have f (x) = P(x) Q(x) = S(x) |{z} polynomial + R(x) Q(x) | {z } proper rational ➤ Step 2 is to factor Q(x) as far as possible.
- 16. Factor Q(x) ➤ Theorem: Any polynomial Q(x) can be factored as a product of ➤ linear factors ax + b ➤ and irreducible quadratic factors ax2 + bx + c where b2 − 4ac < 0 ➤ For example, x4 − 16 = (x2 + 4)(x2 − 4) = (x2 + 4) | {z } irreducible (x + 2) | {z } linear (x − 2) | {z } linear
- 18. Partial Fraction Decomposition ➤ After Step 2 we have f (x) = S(x) |{z} polynomial + R(x) Q(x) | {z } proper rational where Q is completely factored into its I) distinct linear factors II) repeated linear factors III) distinct irreducible quadratic factors IV) repeated irreducible quadratic factors ➤ Q(x) may have any combination of these four types of factors. ➤ Step 3 is the partial fraction decomposition of R(x) Q(x).
- 19. Partial Fractions ➤ Theorem: Any proper rational function R(x) Q(x) can be written as the sum of partial fractions of the form A (ax + b)i or Ax + B (ax2 + bx + c)j where i, j are positive integers. ➤ The types of factors present in Q(x) (I,II, III, IV) will determine the partial fraction decomposition. ➤ The coefficients A, B, and C must be computed algebraically.
- 20. Case I: Distinct Linear Factors
- 21. Case I: Q(x) is a Product of Distinct Linear Factors ➤ Case I: Q(x) can be written as a product of distinct linear factors Q(x) = (a1x + b1)(a2x + b2) . . . (akx + bk) where no factor is repeated and no factor is a constant multiple of another. ➤ In this case the theorem of partial fractions states that there exist constants A1, A2, . . . , Ak such that R(x) Q(x) = A1 a1x + b1 + A2 a2x + b2 + . . . + Ak akx + bk (1)
- 22. Example 2 <1/7> Evaluate Z x2 + 2x − 1 2x3 + 3x2 − 2x dx ➤ Step 1: This rational function is proper since the degree of the numerator is 2 and the degree of the denominator is 3. This means we do not have to do long division. ➤ Step 2: Factor the denominator: x2 + 2x − 1 2x3 + 3x2 − 2x = x2 + 2x − 1 x(2x2 + 3x − 2) = x2 + 2x − 1 x(2x − 1)(x + 2) The denominator factors into 3 distinct linear factors.
- 23. Example 2 <2/7> ➤ Step 3: Partial fraction decomposition: ➤ Step 3a: x2 + 2x − 1 x(2x − 1)(x + 2) = A1 x + A2 2x − 1 + A3 x + 2 ➤ Step 3b: Solve for the unknown constants A1, A2, and A3 by first multiplying both sides of the equation by Q(x): x2 + 2x − 1 = A1 x(2x − 1)(x + 2) x + A2x (2x − 1)(x + 2) 2x − 1 . . . . . . + A3x(2x − 1) (x + 2) x + 2 = A1(2x − 1)(x + 2) + A2x(x + 2) + A3x(2x − 1) ➤ Q(x) obviously cancels out on the left side, and on the right side a factor cancels with the denominator for each term.
- 24. Example 2 3/7 ➤ Now we have the equation x2 +2x−1 = A1(2x−1)(x+2)+A2x(x+2)+A3x(2x−1) (♠) ➤ There are two different methods we will outline for how to find the constants from this point. ➤ Step 3c; Method 1: Distribute and match: x2 + 2x − 1 = A1(2x2 + 3x − 2) + A2(x2 + 2x) + A3(2x2 − x) = (2A1 + A2 + 2A3)x2 + (3A1 + 2A2 − A3)x − 2A1 ➤ First we distributed the factors on right hand side (RHS) of (♠) and collected like terms. ➤ Now we will match the coefficients of xn on each side of the equation (next slide).
- 25. Example 2 4/7 1x2 + 2x−1 = (2A1 + A2 + 2A3)x2 + (3A1 + 2A2 − A3)x + −2A1 2A1 + A2 + 2A3 = 1 (i) 3A1 + 2A2 − A3 = 2 (ii) −2A1 = −1 (iii) ➤ Matching the coefficients we have formed 3 equations for our 3 unknowns A1, A2, A3. This is called a system of linear equations. ➤ We can immediately solve (iii) for A1: A1 = 1 2 ➤ We could then plug this into (i) to solve for A2 in terms of A3. 2 1 2 + A2 + 2A3 = 1 =⇒ A2 = −2A3 ➤ Now plug A1 = 1 2 and A2 = −2A3 into (ii) (next slide).
- 26. Example 2 5/7 3A1 + 2A2 − A3 = 2 =⇒ 3 1 2 + 2(−2A3) − A3 = 2 3 2 − 4A3 − A3 = 2 =⇒ −5A3 = 1 2 =⇒ A3 = − 1 10 ➤ Now since A2 = −2A3, A2 = −2 1 10 =⇒ A2 = 1 5 . ➤ So we have found all of the coefficients and we could then plug them in and integrate. ➤ Remark: There are many other ways you could have solved the system of linear equations. Give it a try.
- 27. Example 2 6/7 ➤ Step 3c; Method 2: Find the zeroes of the Q(x), and sub them into (♠) x2 +2x−1 = A1(2x−1)(x+2)+A2x(x+2)+A3x(2x−1) (♠) and solve for A1, A2, and A3. ➤ The zeroes of (x)(2x − 1)(x + 2) are x = 0, 1 2, and − 2 (x = 0) : −1 = −2A1 =⇒ A1 = 1 2 (x = 1 2 ) : 1 4 = A2 1 2 ( 1 2 + 2) = 5A2 4 =⇒ A2 = 1 5 (x = −2) : −1 = A3(−2)(−4 − 1) = 10 =⇒ A3 = − 1 10
- 28. Example 2 7/7 ➤ Remark: Method 2 will always work for Case I: distinct linear factors. It won’t work immediately for other cases, ie, repeated linear (case II) or irreducible quadratic (cases III IV). Method 1 will work in all possible cases. ➤ Once we have found the coefficients, we plug in and integrate. Z x2 + 2x − 1 2x3 + 3x2 − 2x dx = Z 1 2x + 1 5(2x − 1) + 1 10(x + 2) dx = 1 2 Z 1 x dx + 1 5 Z 1 2x − 1 dx | {z } u1=2x−1, du1=2dx dx= du1 2 + 1 10 Z 1 x + 2 dx | {z } u2=x+2, du2=dx = 1 2 ln |x| + 1 10 ln |2x − 1| − 1 10 ln |x + 2| + C
- 29. Example 3 1/4 Show Z dx x2 − a2 = 1 2a ln
- 32. x − a x + a
- 35. + C where a ̸= 0. This formula can be used for future problems. Try this on you own before going ahead for the solution.
- 36. Example 3 2/4 Z dx x2 − a2 ➤ Step 1: The rational function is proper since the numerator has degree 0 and the denominator has degree 2. No need for long division. ➤ Step 2: Factor the denominator: 1 x2 − a2 = 1 (x − a)(x + a) ➤ Step 3: Partial Fraction Decomposition: Case I : 1 (x − a)(x + a) = A1 x − a + A2 x + a Mult. by Q(x) : 1 = A1(x + a) + A2(x − a)
- 37. Example 3 3/4 ➤ Method 2: The zeroes are x = ±a: 1 = A1(x + a) + A2(x − a) =⇒ (x = −a) : 1 = −2aA2 =⇒ A2 = − 1 2a (x = a) : 1 = 2aA1 =⇒ A1 = 1 2a ➤ Plugging in we have 1 x2 − a2 = 1 2a(x − a) − 1 2a(x + a) Z 1 x2 − a2 dx = Z 1 2a(x − a) − 1 2a(x + a) dx = 1 2a Z 1 x − a dx − 1 2a Z 1 x + a dx
- 38. Example 3 4/4 1 2a Z 1 x − a dx | {z } u=x−a du=dx − 1 2a Z 1 x + a dx | {z } u=x+a du=dx = 1 2a ln |x − a| − 1 2a ln |x + a| + C = 1 2a ln
- 41. x − a x + a
- 44. + C ✓
- 45. Case II: Repeated Linear Factors
- 46. Case II: Repeated Linear Factors ➤ Case II: Q(x) contains a product of repeated linear factors. ➤ Suppose the first linear factor is repeated r times, ie, (a1x + b1)r occurs in the decomposition of Q(x). Then the single term A1 a1x+b1 is replaced by the sum A1 (a1x + b) + A2 (a1 + b1x)2 + . . . + Ar (a1x + b1)r (2) ➤ For example, x3 − x + 1 (x + 1)3x2 = A x + 1 + B (x + 1)2 + C (x + 1)3 + D x + E x2
- 47. Example 3 1/5 Find Z x4 − 2x2 + 4x + 1 x3 − x2 − x + 1 dx ➤ Step 1: This rational function is improper since the degree of the numerator is 4 and the degree of the denominator is 3. This means we must do long division before we can do partial fractions.
- 48. Example 3 2/5 x4 − 2x2 + 4x + 1 x3 − x2 − x + 1 = x + 1 + 4x x3 − x2 − x + 1 ➤ Step 2: Factor Q(x): x3 − x2 − x + 1 = (x3 − x) − (x2 − 1) = x(x2 − 1) − (x2 − 1) = (x2 − 1)(x − 1) = (x + 1)(x − 1)(x − 1) = (x + 1)(x − 1)2 ➤ This method of factoring is called factoring by grouping.
- 49. Example 3 3/5 ➤ Step 3: Partial Fraction Decomposition: 4x x3 − x2 − x + 1 = 4x (x − 1)2 | {z } 2×repeated linear · (x + 1) | {z } distinct linear = A x − 1 + B (x − 1)2 + C x + 1 ➤ Multiply both sides by Q(x): 4x = A(x − 1)(x + 1) + B(x + 1) + C(x − 1)2 (♣) ➤ Q(x) has two zeroes x = ±1 ➤ If we plug x = 1 into (♣), we can solve for B 4 = B(2) =⇒ B = 2
- 50. Example 3 4/5 ➤ If we plug the other zero, x = −1, into (♣) we can solve for C −4 = C(−1 − 1)2 =⇒ −4 = 4C =⇒ C = −1 ➤ But we still have to find A and the only zeros are x = ±1. ➤ No worries, we can plug B and C into (♣) to get the equation 4x = A(x − 1)(x + 1) + 2(x + 1) − (x − 1)2 then plug in any other value for x and then solve for A: (x = 0) : 0 = A(−1)(1) + 2(1) − (−1)2 =⇒ 0 = −A + 2 − 1 =⇒ A = 1 ➤ Remark: Method 1 would have also worked. Notice that, as mentioned earlier, method 2 doesn’t work immediately since we have only 2 zeroes for 3 unknowns. But we can always plug in other values of x to find the remaining unknown(s).
- 51. Example 3 5/5 Z x4 − 2x2 + 4x + 1 x3 − x2 − x + 1 dx = Z x + 1 + 1 x − 1 + 2 (x − 1)2 − 1 x + 1 dx = Z x + 1dx + Z 1 x − 1 dx | {z } u1=x−1 du1=dx +2 Z 1 (x − 1)2 dx | {z } u2=x−1 du2=dx − Z 1 x + 1 dx | {z } u3=x+1 du3=dx = x2 + x + Z 1 u1 du1 + 2 Z 1 u2 2 du2 − Z 1 u3 du3 = x2 + x + ln |u1| − 2 u2 − ln |u3| + C = x2 + x + ln |x − 1| − 2 x − 1 − ln |x + 1| + C = x2 + x + ln
- 54. x − 1 x + 1
- 57. − 2 x − 1 + C
- 58. Case III: Distinct Irreducible Quadratic Factors
- 59. Case III: Distinct Irreducible Quadratic ➤ CASE III: Q(x) contains irreducible quadratic factors, none of which are repeated. If Q(x) has the factor ax2 + bx + c where b2 − 4ac 0, then, in addition to the partial fractions in equations (1) and (2), the expression for R(x) Q(x) will have a term of the form Ax + B ax2 + bx + c (3) For example, x (x − 2)(x2 + 1)(x2 + 4) = A x − 2 + Bx + C x2 + 1 + Dx + E x2 + 4
- 60. Integrating Term (3) ➤ To integrate the term Ax + B ax2 + bx + c (3) · if b = 0 Z Ax + B ax2 + c dx = Z Ax ax2 + c dx | {z } (i) u sub + Z B ax2 + c dx | {z } (ii) use † Z dx x2 + a2 = 1 a tan−1 x a + C (†) · if b ̸= 0, start by completing the square: ax2 + bx + c = a x + b 2a 2 + c − b2 4a then do a u sub, u = x + b 2a , du = dx.
- 61. Example 4 1/4 Evaluate Z 2x2 − x + 4 x3 + 4x dx ➤ Step 1: The rational function is proper so we don’t need to do long division. ➤ Step 2: Factor Q(x): x3 + 4x = x(x2 + 4) = x |{z} distinct linear (x2 + 4) | {z } distinct irreducible quadratic ➤ Remark: Quadratics of the form ax2 + c where a, c 0 are always irreducible since b2 − 4ac = 02 − 4ac = −4ac 0
- 62. Example 4 2/4 2x2 − x + 4 x3 + 4x = A x + Bx + C x2 + 4 ➤ Now multiplying both sides by Q(x) we have 2x2 − x + 4 = A(x2 + 4) + (Bx + C)x (♢) ➤ Since x2 + 4 is irreducible, it has no roots. So Q(x) only has one zero at x = 0. ➤ Plugging x = 0 into ♢ we can solve for A: 4 = 4A =⇒ A = 1 ➤ We can then plug A = 1 into ♢ to get 2x2 − x + 4 = x2 + 4 + (Bx + C)x (♢♢)
- 63. Example 4 3/4 ➤ We can find B and C by method 1 (distribute and match) on ♢♢: 2x2 − x + 4 = x2 + 4 + (Bx + C)x = x2 + 4 + Bx2 + Cx = (B + 1)x2 + Cx + 4 2x2 −1x + 4 = (B + 1)x2 + Cx + 4 B + 1 = 2 =⇒ B = 1 C = −1 ➤ Remark: We could have skipped plugging in x = 0 just used method 1 on ♢. Alternatively, we could’ve found A as we did, then plugged two other values for x, eg x = 12, into ♢♢ to get a system of linear equations to solve for C and B.
- 64. Example 4 4/4 ➤ Plugging in A, B, and C: Z 2x2 − x + 4 x3 + 4x dx = Z 1 x dx + Z x − 1 x2 + 4 dx = ln |x| + Z x x2 + 4 dx | {z } u=x2+4 du=2xdx du 2 =xdx − Z 1 x2 + 4 dx | {z } use †, a=2 = ln |x| + Z du 2u − 1 2 · 2 tan−1 x 2 = ln |x| + 1 2 ln |x2 + 4| − 1 4 tan−1 x 2 + C
- 65. Case IV: Repeated Irreducible Quadratic Factors
- 66. Case IV: Repeated Irreducible Quadratic ➤ Case IV: Q(x) contains a repeated irreducible quadratic factor. ➤ If Q(x) has the factor (ax2 + bx + c)r where b2 − 4ac 0, then, instead of the single partial fraction (3), the sum A1x + B1 ax2 + bx + c + A2x + B3 (ax2 + bx + c)2 +. . .+ Ar x + Br (ax2 + bx + c)r (4) occurs in the partial fraction decomposition of R(x) Q(x). ➤ Each term in (4) can be integrated using substitution, completing the square first if necessary.
- 67. Example 5 1/2 Write the partial fraction decomposition of the following rational function; don’t solve for constants. x3 + x2 + 1 x(x − 2)(x2 + x + 1)(2x2 + 1)3 ➤ x and x − 2 are distinct linear factors which will contribute A x + B x − 2 ➤ x2 + x + 1 has a = 1, b = 1, c = 1 so b2 − 4ac = −3 0. This is a distinct irreducible quadratic which will contribute Cx + D x2 + x + 1 ➤ (2x2 + 1)3 is an irreducible quadratic repeated (×3). It will contribute Ex + F 2x2 + 1 + Gx + H (2x2 + 1)2 + Ix + J (2x2 + 1)3
- 68. Example 5 2/2 So the partial fraction decomposition of x3 + x2 + 1 x(x − 2)(x2 + x + 1)(2x2 + 1)3 is A x + B x − 2 + Cx + D x2 + x + 1 + Ex + F 2x2 + 1 + Gx + H (2x2 + 1)2 + Ix + J (2x2 + 1)3
- 69. Example 6 1/6 Evaluate Z 1 − x + 2x2 − x3 x(x2 + 1)2 dx ➤ Step 1: The numerator is of degree 3 and the denominator is degree 1 + 2 · 2 = 5 so this is proper. ➤ Step 2: Since x2 + 1 is obviously irreducible, Q(x) is already completely factored. ➤ Step 3: Partial fraction decomposition: 1 − x + 2x2 − x3 x(x2 + 1)2 = A x + Bx + C x2 + 1 + Dx + E (x2 + 1)2 ▶ Multiplying both sides by Q(x): 1−x +2x2 −x3 = A(x2 +1)2 +(Bx +C)x(x2 +1)+(Dx +E)x (♡)
- 70. Example 6 2/6 ➤ Again, Q(x) only has one root, x = 0. Plugging this into ♡ we get A = 1 and plugging A = 1 into ♡ we get 1−x +2x2 −x3 = (x2 +1)2 +(Bx +C)x(x2 +1)+(Dx +E)x (♡♡) ➤ Now we use method 1 on ♡♡: distribute and match: 1 − x + 2x2 − x3 = (x4 + 2x2 + 1) + (Bx2 + Cx)(x2 + 1) + (Dx2 + Ex) 1 − x + 2x2 − x3 = (x4 + 2x2 + 1) + (Bx2 + Cx)(x2 + 1) + (Dx2 + Ex) −x − x3 = x4 + (Bx4 + Bx2 + Cx3 + Cx) + (Dx2 + Ex) = (1 + B)x4 + Cx3 + (B + D)x2 + (C + E)x
- 71. Example 6 3/6 0x4 −1x3 + 0x2 −1x = (1 + B)x4 + Cx3 + (B + D)x2 + (C + E)x 0 = 1 + B =⇒ B = −1 C = −1 0 = B + D =⇒ D = −B =⇒ D = 1 −1 = C + E =⇒ E = −1 − C =⇒ E = 0 1 − x + 2x2 − x3 x(x2 + 1)2 = A x + Bx + C x2 + 1 + Dx + E (x2 + 1)2 = 1 x + −x − 1 x2 + 1 + x (x2 + 1)2
- 72. Example 6 4/6 Z 1 x + −x − 1 x2 + 1 + x (x2 + 1)2 dx = Z dx x | {z } (i) − Z x + 1 x2 + 1 dx | {z } (ii) + Z x (x2 + 1)2 dx | {z } (iii) (i) Z dx x = ln |x| + C1 (ii) Z x + 1 x2 + 1 dx = Z x x2 + 1 dx | {z } u=x2+1 du=2xdx du 2 =dx + Z 1 x2 + 1 dx | {z } use †, a=1 = Z du 2u + tan−1 x = 1 2 ln |x2 + 1| + tan−1 x + C2
- 73. Example 6 5/6 (iii) Z x − 2 (x2 + 1)2 dx = Z x (x2 + 1)2 dx | {z } u=x2+1 du=2x du 2 =xdx = Z 1 2u2 du = − 1 2(x2 + 1) + C3
- 74. Example 6 6/6 Putting it all together we have: ln |x| − 1 2 ln |x2 + 1| + tan−1 x − 1 2(x2 + 1) + C where C = C1 + C2 + C3. We could just wait until the very end to add C of course.
- 76. Rationalizing Substitution ➤ When an integrand contains an expression of the form n p g(x) then the substitution u = n p g(x) may be helpful. ➤ For example, to evaluate Z √ x + 4 x dx we make the substitution u = √ x + 4. ➤ Then to compute dx in terms of u and du we can · Solve for x and then differentiate: u2 = x + 4 =⇒ x = u2 − 4 =⇒ dx = 2udu · Or do implicit differentiation: du dx = d dx √ x + 4 = 1 2 √ x + 4 = 1 2u =⇒ dx = 2udu
- 77. Z √ x + 4 x dx = Z u u2 − 4 2udu = 2 Z u2 u2 − 4 du = 2 Z u2 − 4 + 4 u2 − 4 du = 2 Z u2 − 4 u2 − 4 + 4 u2 − 4 du = 2 Z 1 + 4 u2 − 4 du = 2u + 8 Z du u2 − 4 = 2u + 8 1 2 · 2 ln
- 80. u − 2 u + 2
- 83. + C = 2 √ x + 4 + 2 ln
- 86. √ x + 4 − 2 √ x + 4 + 2
- 89. + C