anonymous 5 years ago integrate by partial fractions, not sure,. i want to integrate 1/((x^2*(x^2-15)) should i do partial fraction 1/((x^2*(x^2-15)) = A / x + B/x^2 + C/(x-sqrt 15) + D/(x+sqrt 15) , or 1/((x^2*(x^2-15)) = A / x + B/x^2 + Cx+d / ( x^2 - 15)

1. amistre64

oy... if you hit the enter key it spreads this out to where it is more readable.... just a thought

2. anonymous

the x^2 - 15 is reducible , but im not sure if youre supposed to use that theorem

3. anonymous

enter key?

4. amistre64

the key on the keyboard that makes this happen ;)

5. anonymous

you want me to use the enter key more often?

6. amistre64

yes, difference of squares that part and you should have 3 sets to figure out

7. anonymous

8. amistre64

9. anonymous

are they ? how come the other way works just as well?

10. amistre64

which other way can you decompose the fraction?

11. anonymous

1/((x^2*(x^2-15)) = A / x + B/x^2 + (Cx+d) / ( x^2 - 15)

12. anonymous

13. amistre64

it could be a fluke if it does work out....

14. amistre64

or it could be a possibility for solutions that i am unaware of as yet :)

15. anonymous

no it isnt a fluke, in general one need not reduce a quadratic

16. amistre64

in general you should, to account for multiples i think

17. amistre64

if it aint got multiples, then you should be fine i spose

18. amistre64

right?

19. anonymous

yes it usually makes it easier to reduce the quadratic, except in the case where there are radical linear factors

20. anonymous

it can have multiples, theres no problem as far as i can see

21. amistre64

without doing a ton of them to generalize it, id have to throw it in the hmmmmm box lol

22. anonymous

say integral ( 1 / (x(x+1)^2) = A/x + (Bx+C) / (x+1)^2

23. anonymous

its a little weird, might be easier if i wrote 1/( x ( x^2+2x+1)) = A/x + (Bx+c) / (x^2 + 2x + 1)

24. anonymous

sometimes it is difficult to factor a quadratic, right?

25. anonymous

ok , i should get into more accurate terms. sometimes it is reducible over reals, not rationals

26. amistre64

factoring a quadratic is fairly easy; might get complicated solutions, but farily easy to do

27. anonymous

yes but doing partial fractions with irrational roots is tough

28. amistre64

and i have heard that you can use the complex solutions as well with no ill effects on the decomp

29. anonymous

ok , so youre so confident, try to do parfractions on 1 / ( x^2(x^2-15) , give it a go

30. anonymous

i was stuck

31. anonymous

1/((x^2*(x^2-15)) = A / x + B/x^2 + C/(x-sqrt 15) + D/(x+sqrt 15)

32. amistre64

confident? lol ; just stubborn

33. anonymous

it can even get stranger, this is an easy irrational solution problem, you can have say A/ ( x - sqrt 3/2 ) for instance

34. anonymous

this is why calculus in textbooks is generally contrived , or idealized problems. in the real world you have these irrational roots, etc

35. anonymous

theorem time, one sec

36. amistre64

1 a b c d ----------- = --- + ----- + --------- + --------- x^2(x^2-15) x x^2 x-sqrt(15) x+sqrt(15) is the 'proper' set up right?

37. anonymous

yes looks good

38. anonymous

Personally, I'd leave the last fraction's denominator without doing difference of two squares.

39. anonymous

1 / [(x-a)^n (x^2+bx+c)^n] = A1/(x-a) + A2/ ( x-a)^2 + ... An/(x-a)^n + (B1x+C1)/ (x^2+bx+c) + (B2x + C2) / (x^2+bx+c)^2 + ... + (Bn x + Cn ) / (x^2+bx+c)^n

40. anonymous

so the question is, do the quadratics have to be irreducible, i dont think so

41. amistre64

1 a b c d ----------- = --- + ----- + --------- + --------- x^2(x^2-15) x x^2 x-sqrt(15) x+sqrt(15) a(x^2)(x-sqrt(15))(x+sqrt(15))) b(x)(x-sqrt(15))(x+sqrt(15))) c(x)( x^2)(x+sqrt(15)) d(x)(x^2)(x-sqrt(15))

42. anonymous

MOST calculus books and websites say the quadratic is irreducible though

43. anonymous

irreducible over the reals i imagine. wikipedia is better, they say irreducible over the field

44. anonymous

whatever field you have in mine

45. anonymous

For this particular problem: $\frac{1}{y(y-15)} \equiv \frac{1}{15(y-15)} - \frac{1}{15y}$ $\text{Let } y = x^2$ ...

46. anonymous

newton, be careful though

47. anonymous

if you are substituting a variable, you will need to do 2x dx = dy

48. anonymous

so you might want to just change it back. thats an identity

49. anonymous

i should write a book on calculus pitfalls, since i keep finding them :)

50. anonymous

newtons, so just change y back to x^2

51. anonymous

Sorry, in case it wasn't clear, it leads to: $\frac{1}{x^2(x^2-15)} \equiv \frac{1}{15(x^2-15)} + \frac{1}{15x^2}$ I would integrate this.

52. anonymous

otherwise you have 2x dx = dy , dx = dy / ( 2x ) = dy / ( 2 sqrt y )

53. anonymous

correct, i thought you were doing a change of variables

54. amistre64

1 = a(x^2)(x-sqrt(15))(x+sqrt(15))) + b(x)(x-sqrt(15))(x+sqrt(15))) + c(x)( x^2)(x+sqrt(15)) +d(x)(x^2)(x-sqrt(15)) when x = sqrt(15) we get 1 = c(sqrt(15))(15)(2sqrt(15)) 1 = c(15)(2)(15) 1 = c(225)(2) = c450 c = 1/450

55. anonymous

amistre, you get medal for persistence :)

56. anonymous

but its much easier if we dont split the quadratic

57. anonymous

I was only doing that to show that you don't need a 1/x fraction and 1/x^2 , as x^2 is the only term. I wasn't integrating with a different variable.

58. amistre64

yay!! lol :) can I stop now lol

59. anonymous

ok

60. anonymous

wait that doesnt work though

61. anonymous

I mixed up a minus and plus sign ¬_¬

62. anonymous

its incorrect to do 1 / ( x^2 (x^2-15 ) = A / x^2 + b / (x^2-15) , thats incorrect

63. anonymous

you wont be able to find A and B

64. anonymous

I just did?

65. anonymous

you used a simpler case though

66. anonymous

Multiply it out, plz.

67. anonymous

1 = A(x^2 - 15) + B x^2

68. anonymous

, 1 = Ax^2 + bx^2 - 15

69. anonymous

(the + the last fraction should be minus, as in the first time, typo)

70. anonymous

1 = Ax^2 + Bx62 - 15 A

71. anonymous

hmmmm,

72. anonymous

but the book says it should be A/ x + B/x^2 + C x+d / (x^2 - 15)

73. anonymous

well this approach works as well

74. anonymous

You can do it if you want, but you'll find A = 0, so it doesn't really matter.

75. anonymous

right, so come the theorem doesnt work in these websites

76. anonymous

as far as repeated roots go

77. anonymous

for instance you didnt use CX+D, i think its because of that y = x^2 business,

78. anonymous

There are only terms in x^2 in this case, it wouldn't always work. The general method is probably more ... general.

79. anonymous

ahh

80. anonymous

so my question, my original question, hmmmm, do the quadratics have to be irreducible in the general case

81. anonymous

ok for instance, 1 / (( x-2)^2 ( x^2-15))

82. anonymous

1 / (( x-2)^2 ( x^2-15)) = A / (x-2) + b / (x-2)^2 + (cx+d) / (x^2 - 15 ) ?

83. anonymous

I'd say it's probably easier to leave it as the irreducible (without using difference of two squares) quadratic with Cx + d. I guess you could split up this factor, but it's unnecessary (as far as I know)

84. anonymous

Yes, that.

85. anonymous

but its not technically irreducible

86. anonymous

i can even go so far as to do this

87. anonymous

1 / (( x-2)^2 ( x^2-15)) = (Ax+b) / (x-2)^2 + (cx+d) / (x^2 - 15 )

88. anonymous

no chat room

89. anonymous

sorry, not for you

90. anonymous

so this whole irreducible business is unnecessary

91. anonymous

1 = a*x^3-15*a*x+b*x^2-15*b+c*x^3-4*c*x^2+4*c*x+d*x^2-4*d*x+4*d

92. anonymous

ok im going to solve an easier one

93. anonymous

int 1/ ((x-2)(x^2-15))

94. anonymous

1/ ((x-2)(x^2-15)) = A/ (x-2) + (Bx+c) / ( x^ 2 - 15)