17 integrals of rational functions x

Integrals of Rational Functions

1. In this section we will show how to integrate rational functions, that is, functions of the form P(x) Q(x) where P and Q are polynomials. Integrals of Rational Functions

2. In this section we will show how to integrate rational functions, that is, functions of the form P(x) Q(x) where P and Q are polynomials. Rational Decomposition Theorem Given reduced P/Q where deg P < deg Q, then P/Q = F 1 + F 2 + .. + F n where F i = or A (ax + b) k Ax + B (ax 2 + bx + c) k and that (ax + b) k or (ax 2 + bx + c) k are factors of Q(x). Integrals of Rational Functions

3. Therefore finding the integrals of rational functions is reduced to finding integrals of or A (ax + b) k Ax + B (ax 2 + bx + c) k Integrals of Rational Functions

4. Therefore finding the integrals of rational functions is reduced to finding integrals of or A (ax + b) k Ax + B (ax 2 + bx + c) k Integrals of Rational Functions The integrals ∫ are straight forward with the substitution method by setting u = ax + b. dx (ax + b) k

5. Therefore finding the integrals of rational functions is reduced to finding integrals of or A (ax + b) k Ax + B (ax 2 + bx + c) k Integrals of Rational Functions The integrals ∫ are straight forward with the substitution method by setting u = ax + b. dx (ax + b) k To integrate ∫ dx , we need to complete the square of the denominator. Ax + B (ax 2 + bx + c) k

6. Integrals of Rational Functions x x 2 + 6x + 10 Example A. Find ∫ dx

7. Complete the square of x 2 + 6x + 10 – it’s irreducible. Integrals of Rational Functions x x 2 + 6x + 10 Example A. Find ∫ dx

8. x 2 + 6x + 10 = (x 2 + 6x ) + 10 Integrals of Rational Functions Complete the square of x 2 + 6x + 10 – it’s irreducible. x x 2 + 6x + 10 Example A. Find ∫ dx

9. x 2 + 6x + 10 = (x 2 + 6x + 9 ) + 10 – 9 Integrals of Rational Functions Complete the square of x 2 + 6x + 10 – it’s irreducible. x x 2 + 6x + 10 Example A. Find ∫ dx

10. x 2 + 6x + 10 = (x 2 + 6x + 9 ) + 10 – 9 = (x + 3) 2 + 1 Integrals of Rational Functions Complete the square of x 2 + 6x + 10 – it’s irreducible. x x 2 + 6x + 10 Example A. Find ∫ dx

11. x 2 + 6x + 10 = (x 2 + 6x + 9 ) + 10 – 9 = (x + 3) 2 + 1 x x 2 + 6x + 10 ∫ dx = ∫ x (x + 3) 2 + 1 dx Hence Integrals of Rational Functions Complete the square of x 2 + 6x + 10 – it’s irreducible. x x 2 + 6x + 10 Example A. Find ∫ dx

12. x 2 + 6x + 10 = (x 2 + 6x + 9 ) + 10 – 9 = (x + 3) 2 + 1 x x 2 + 6x + 10 ∫ dx Set u = x + 3 = ∫ x (x + 3) 2 + 1 dx substitution Hence Integrals of Rational Functions Complete the square of x 2 + 6x + 10 – it’s irreducible. x x 2 + 6x + 10 Example A. Find ∫ dx

13. x 2 + 6x + 10 = (x 2 + 6x + 9 ) + 10 – 9 = (x + 3) 2 + 1 x x 2 + 6x + 10 ∫ dx Set u = x + 3 x = u – 3 dx = du = ∫ x (x + 3) 2 + 1 dx substitution Hence Integrals of Rational Functions Complete the square of x 2 + 6x + 10 – it’s irreducible. x x 2 + 6x + 10 Example A. Find ∫ dx

14. x 2 + 6x + 10 = (x 2 + 6x + 9 ) + 10 – 9 = (x + 3) 2 + 1 x x 2 + 6x + 10 ∫ dx Set u = x + 3 x = u – 3 dx = du = ∫ x (x + 3) 2 + 1 dx substitution = ∫ u – 3 u 2 + 1 du Hence Integrals of Rational Functions Complete the square of x 2 + 6x + 10 – it’s irreducible. x x 2 + 6x + 10 Example A. Find ∫ dx

15. x 2 + 6x + 10 = (x 2 + 6x + 9 ) + 10 – 9 = (x + 3) 2 + 1 x x 2 + 6x + 10 ∫ dx Set u = x + 3 x = u – 3 dx = du = ∫ x (x + 3) 2 + 1 dx substitution = ∫ u – 3 u 2 + 1 du = ∫ u u 2 + 1 du – 3 ∫ 1 u 2 + 1 du Hence Integrals of Rational Functions Complete the square of x 2 + 6x + 10 – it’s irreducible. x x 2 + 6x + 10 Example A. Find ∫ dx

16. x 2 + 6x + 10 = (x 2 + 6x + 9 ) + 10 – 9 = (x + 3) 2 + 1 x x 2 + 6x + 10 ∫ dx Set u = x + 3 x = u – 3 dx = du = ∫ x (x + 3) 2 + 1 dx substitution = ∫ u – 3 u 2 + 1 du = ∫ u u 2 + 1 du – 3 ∫ 1 u 2 + 1 du Hence Integrals of Rational Functions Set w = u 2 + 1 substitution Complete the square of x 2 + 6x + 10 – it’s irreducible. x x 2 + 6x + 10 Example A. Find ∫ dx

17. Set w = u 2 + 1 = 2u substitution dw du Integrals of Rational Functions = ∫ u u 2 + 1 du – 3 ∫ 1 u 2 + 1 du Set w = u 2 + 1

18. Set w = u 2 + 1 = 2u substitution dw du du = dw 2u Integrals of Rational Functions = ∫ u u 2 + 1 du – 3 ∫ 1 u 2 + 1 du Set w = u 2 + 1

19. Set w = u 2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u 2 + 1 du dw 2u Integrals of Rational Functions = ∫ u u 2 + 1 du – 3 ∫ 1 u 2 + 1 du Set w = u 2 + 1

20. Set w = u 2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u 2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan -1 (u) + c dw Integrals of Rational Functions

21. Set w = u 2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u 2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan -1 (u) + c dw = ½ Ln(w) – 3 tan -1 (x + 3) + c Integrals of Rational Functions

22. Set w = u 2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u 2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan -1 (u) + c dw = ½ Ln( lwl ) – 3 tan -1 (x + 3) + c = ½ Ln((x + 3) 2 + 1) – 3 tan -1 (x + 3) + c Integrals of Rational Functions

23. Set w = u 2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u 2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan -1 (u) + c dw = ½ Ln( lwl ) – 3 tan -1 (x + 3) + c = ½ Ln((x + 3) 2 + 1) – 3 tan -1 (x + 3) + c Integrals of Rational Functions 3x x 2 + 7x + 10 Example: Find ∫ dx

24. Set w = u 2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u 2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan -1 (u) + c dw = ½ Ln( lwl ) – 3 tan -1 (x + 3) + c = ½ Ln((x + 3) 2 + 1) – 3 tan -1 (x + 3) + c Integrals of Rational Functions Since x 2 + 7x + 10 = (x + 2)(x + 5), we can decompose the rational expression. 3x x 2 + 7x + 10 Example: Find ∫ dx

25. Set w = u 2 + 1 = 2u substitution dw du du = dw 2u = ∫ u w – 3 ∫ 1 u 2 + 1 du dw 2u = ½ ∫ 1 w – 3 tan -1 (u ) + c dw = ½ Ln( lwl ) – 3 tan -1 (x + 3) + c = ½ Ln((x + 3) 2 + 1) – 3 tan -1 (x + 3) + c Integrals of Rational Functions 3x x 2 + 7x + 10 Example: Find ∫ dx Since x 2 + 7x + 10 = (x + 2)(x + 5), we can decompose the rational expression. Specifically 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)

26. Integrals of Rational Functions Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)

27. Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)

28. Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)

29. Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)

30. Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)

31. Integrals of Rational Functions 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)

32. Integrals of Rational Functions Hence 3x (x + 2)(x + 5) = -2 (x + 2) + 5 (x + 5) 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)

33. Integrals of Rational Functions Hence x (x + 2)(x + 5) = -2 (x + 2) + 5 (x + 5) 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 3x x 2 + 7x + 10 So ∫ dx Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)

34. Integrals of Rational Functions Hence x (x + 2)(x + 5) = -2 (x + 2) + 5 (x + 5) 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 3x x 2 + 7x + 10 So ∫ dx = ∫ dx -2 (x + 2) + 5 (x + 5) Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)

35. Integrals of Rational Functions Hence x (x + 2)(x + 5) = -2 (x + 2) + 5 (x + 5) 3x = A(x + 5) + B(x + 2) Evaluate at x = -5, we get -15 = -3B so B = 5 Evaluate at x = -2, we get -6 = 3A so A = -2 3x x 2 + 7x + 10 So ∫ dx = - 2Ln(lx + 2l) + 5Ln(lx + 5l) + c = ∫ dx -2 (x + 2) + 5 (x + 5) Clear denominators: 3x (x + 2)(x + 5) = A (x + 2) + B (x + 5)

36. Integrals of Rational Functions 1 (x 2 + 1) 2 x Example: Find ∫ dx

37. Integrals of Rational Functions 1 (x 2 + 1) 2 x = Ax + B (x 2 + 1) + Cx + D (x 2 + 1) 2 + E x 1 (x 2 + 1) 2 x Example: Find ∫ dx

38. Integrals of Rational Functions 1 (x 2 + 1) 2 x Example: Find ∫ dx 1 (x 2 + 1) 2 x = Ax + B (x 2 + 1) + Cx + D (x 2 + 1) 2 + E x Clear denominators

39. Integrals of Rational Functions 1 (x 2 + 1) 2 x = Ax + B (x 2 + 1) + Cx + D (x 2 + 1) 2 + E x Clear denominators 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + E (x 2 + 1) 2 1 (x 2 + 1) 2 x Example: Find ∫ dx

40. Integrals of Rational Functions 1 (x 2 + 1) 2 x = Ax + B (x 2 + 1) + Cx + D (x 2 + 1) 2 + E x Clear denominators 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + E (x 2 + 1) 2 Evaluate this at x = 0, we get 1 = E 1 (x 2 + 1) 2 x Example: Find ∫ dx

41. Integrals of Rational Functions 1 (x 2 + 1) 2 x = Ax + B (x 2 + 1) + Cx + D (x 2 + 1) 2 + E x Clear denominators 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + E (x 2 + 1) 2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 1 (x 2 + 1) 2 x Example: Find ∫ dx

42. Integrals of Rational Functions 1 (x 2 + 1) 2 x = Ax + B (x 2 + 1) + Cx + D (x 2 + 1) 2 + E x Clear denominators 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + E (x 2 + 1) 2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 There is no x 4 -term, 1 (x 2 + 1) 2 x Example: Find ∫ dx

43. Integrals of Rational Functions 1 (x 2 + 1) 2 x = Ax + B (x 2 + 1) + Cx + D (x 2 + 1) 2 + E x Clear denominators 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + E (x 2 + 1) 2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 There is no x 4 -term, hence Ax 4 + x 4 = 0, 1 (x 2 + 1) 2 x Example: Find ∫ dx

44. Integrals of Rational Functions 1 (x 2 + 1) 2 x = Ax + B (x 2 + 1) + Cx + D (x 2 + 1) 2 + E x Clear denominators 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + E (x 2 + 1) 2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 There is no x 4 -term, hence Ax 4 + x 4 = 0, so A = -1 1 (x 2 + 1) 2 x Example: Find ∫ dx

45. Integrals of Rational Functions 1 (x 2 + 1) 2 x = Ax + B (x 2 + 1) + Cx + D (x 2 + 1) 2 + E x Clear denominators 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + E (x 2 + 1) 2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 There is no x 4 -term, hence Ax 4 + x 4 = 0, so A = -1 There is no x 3 -term, 1 (x 2 + 1) 2 x Example: Find ∫ dx

46. Integrals of Rational Functions 1 (x 2 + 1) 2 x = Ax + B (x 2 + 1) + Cx + D (x 2 + 1) 2 + E x Clear denominators 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + E (x 2 + 1) 2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 There is no x 4 -term, hence Ax 4 + x 4 = 0, so A = -1 There is no x 3 -term, hence Bx 3 = 0, 1 (x 2 + 1) 2 x Example: Find ∫ dx

47. Integrals of Rational Functions 1 (x 2 + 1) 2 x = Ax + B (x 2 + 1) + Cx + D (x 2 + 1) 2 + E x Clear denominators 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + E (x 2 + 1) 2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 There is no x 4 -term, hence Ax 4 + x 4 = 0, so A = -1 There is no x 3 -term, hence Bx 3 = 0, so B = 0 1 (x 2 + 1) 2 x Example: Find ∫ dx

48. Integrals of Rational Functions 1 (x 2 + 1) 2 x = Ax + B (x 2 + 1) + Cx + D (x 2 + 1) 2 + E x Clear denominators 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + E (x 2 + 1) 2 Evaluate this at x = 0, we get 1 = E Hence 1 = (Ax + B) (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 There is no x 4 -term, hence Ax 4 + x 4 = 0, so A = -1 There is no x 3 -term, hence Bx 3 = 0, so B = 0 So we've 1 = -x (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 1 (x 2 + 1) 2 x Example: Find ∫ dx

49. Integrals of Rational Functions There is no x-term in 1 = -x (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2

50. Integrals of Rational Functions There is no x-term in 1 = -x (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 Hence Dx = 0, or D = 0.

51. Integrals of Rational Functions There is no x-term in 1 = -x (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 Hence Dx = 0, or D = 0. So the expression is 1 = -x 2 (x 2 + 1) + Cx 2 + 1 (x 2 + 1) 2

52. Integrals of Rational Functions There is no x-term in 1 = -x (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 Hence Dx = 0, or D = 0. So the expression is 1 = -x 2 (x 2 + 1) + Cx 2 + 1 (x 2 + 1) 2 Expand this

53. Integrals of Rational Functions There is no x-term in 1 = -x (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 Hence Dx = 0, or D = 0. So the expression is 1 = -x 2 (x 2 + 1) + Cx 2 + 1 (x 2 + 1) 2 Expand this 1 = -x 4 – x 2 Cx 2 + x 4 + 2x 2 + 1 +

54. Integrals of Rational Functions There is no x-term in 1 = -x (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 Hence Dx = 0, or D = 0. So the expression is 1 = -x 2 (x 2 + 1) + Cx 2 + 1 (x 2 + 1) 2 Expand this 1 = -x 4 – x 2 Cx 2 + x 4 + 2x 2 + 1 + Hence Cx 2 + x 2 = 0 or C = -1

55. Integrals of Rational Functions There is no x-term in 1 = -x (x 2 + 1) x + (Cx + D) x + 1 (x 2 + 1) 2 Hence Dx = 0, or D = 0. So the expression is 1 = -x 2 (x 2 + 1) + Cx 2 + 1 (x 2 + 1) 2 Expand this 1 = -x 4 – x 2 Cx 2 + x 4 + 2x 2 + 1 + Hence Cx 2 + x 2 = 0 or C = -1 Put it all together 1 (x 2 + 1) 2 x = -x (x 2 + 1) + -x (x 2 + 1) 2 + 1 x

56. Integrals of Rational Functions 1 (x 2 + 1) 2 x Therefore ∫ dx = ∫ -x dx (x 2 + 1) + -x dx (x 2 + 1) 2 + dx x ∫ ∫

57. Integrals of Rational Functions 1 (x 2 + 1) 2 x Therefore ∫ dx = ∫ -x dx (x 2 + 1) + -x dx (x 2 + 1) 2 + dx x ∫ ∫ substitution set u = x 2 + 1

58. Integrals of Rational Functions 1 (x 2 + 1) 2 x Therefore ∫ dx = ∫ -x dx (x 2 + 1) + -x dx (x 2 + 1) 2 + dx x ∫ ∫ substitution set u = x 2 + 1 du dx = 2x

59. Integrals of Rational Functions 1 (x 2 + 1) 2 x Therefore ∫ dx = ∫ -x dx (x 2 + 1) + -x dx (x 2 + 1) 2 + dx x ∫ ∫ substitution set u = x 2 + 1 du dx = 2x du 2x dx =

60. Integrals of Rational Functions absolute value arguments for Ln: P - 22,36,61-64

61. Integrals of Rational Functions 1 (x 2 + 1) 2 x Therefore ∫ dx = ∫ -x dx (x 2 + 1) + -x dx (x 2 + 1) 2 + dx x ∫ ∫ substitution set u = x 2 + 1 du dx = 2x du 2x dx = = ∫ -x u du 2x + ∫ -x u 2 du 2x + Ln( lxl )

62. Integrals of Rational Functions 1 (x 2 + 1) 2 x Therefore ∫ dx = ∫ -x dx (x 2 + 1) + -x dx (x 2 + 1) 2 + dx x ∫ ∫ substitution set u = x 2 + 1 du dx = 2x du 2x dx = = ∫ -x u du 2x + ∫ -x u 2 du 2x + Ln( lxl ) = - ½ ∫ du u + Ln( lxl ) ½ ∫ du u 2 –

63. Integrals of Rational Functions 1 (x 2 + 1) 2 x Therefore ∫ dx = ∫ -x dx (x 2 + 1) + -x dx (x 2 + 1) 2 + dx x ∫ ∫ substitution set u = x 2 + 1 du dx = 2x du 2x dx = = ∫ -x u du 2x + ∫ -x u 2 du 2x + Ln(x) = - ½ ∫ du u + Ln(x) ½ ∫ du u 2 – = - ½ Ln(u) + ½ u -1 + Ln(x) + c

64. Integrals of Rational Functions 1 (x 2 + 1) 2 x Therefore ∫ dx = ∫ -x dx (x 2 + 1) + -x dx (x 2 + 1) 2 + dx x ∫ ∫ substitution set u = x 2 + 1 du dx = 2x du 2x dx = = ∫ -x u du 2x + ∫ -x u 2 du 2x + Ln(x) = - ½ ∫ du u + Ln(x) ½ ∫ du u 2 – = - ½ Ln(u) + ½ u -1 + Ln(x) + c = - ½ Ln(x 2 + 1) + + Ln(x) + c 1 2(x 2 + 1)