## anonymous 3 years ago Integral by partial fraction?

1. anonymous

|dw:1358736744745:dw|

2. abb0t

$4x = \frac{ A }{ (x+1) } + \frac{ Bx+C }{ (x^2+1) } + \frac{ Dx+E }{ (x^2+1)^2 }$

3. anonymous

thank you that helps alot.. how do you come up with that?

4. anonymous

and why is it 4x = ......

5. anonymous

if it is quadratic in denominator in is linear in numerator?

6. abb0t

4x is the numerator. and when it's quadratic it follows the form Ax+B

7. anonymous

so now we solve all the a,b,c,d,e?

8. anonymous

how do we solve for all those?

9. abb0t

You get rid of the denominators.

10. abb0t

$4x = A(x^2+1)^2(x+1) + Bx+C(x+1)(x^2+1) + Dx+E(x+1)$

11. zepdrix

$\large \color{orangered}{\frac{4x}{(x^2+1)^2(x+1)}} = \frac{ A }{ (x+1) } + \frac{ Bx+C }{ (x^2+1) } + \frac{ Dx+E }{ (x^2+1)^2 }$ The left should look like this before you get rid of the denominators :) The rest of that setup looks great though.

12. anonymous

ahh, thank you zepdrix, was a little confused

13. anonymous

I think i can handle it from here, I just have hard time setting the partial fractions up. any tips?

14. abb0t

Give me a minute to write it all out for you, how in general it works

15. anonymous

take your time, i'll work on this problem, thanks so much

16. anonymous

$4x = A{\left( {{x^2} + 1} \right)^2} + Bx + C\left( {{x^2} + 1} \right)\left( {x + 1} \right) + Dx + E\left( {x + 1} \right)$

17. zepdrix

When you multiply through by that denominator you should get, \large \begin{align*} 4x = &A(x^2+1)^2\\&+(Bx+C)(x^2+1)(x+1)\\&+(Dx+E)(x+1) \end{align*} There may have been a small boo boo when abbot multiplied that out, so I just wanted to check.

18. zepdrix

No I guess it was fine, just don't forget the brackets on top.

19. zepdrix

You can easily solve for A by plugging in x=-1. That should give you a 0 on the last 2 large terms. For the rest of them... you have to deal with a nasty squared X term. So you'll have to expand aaaaaaall of that out :( kinda sucks...

20. anonymous

@abb0t did have an extra (x+1) on the first partial..

21. zepdrix

Oh i see c:

22. abb0t

FOR: $ax+b$ the term in partial fraction decomposition is: $\frac{ A }{ ax+b }$ FOR: $(ax+b)^n$ the term in partial fraction decomposition is: $\frac{ A_1 }{ ax+b }+\frac{ A_2 }{ (ax+b)^2 }+...+\frac{ A_n }{ (ax+b)^n }$ where n = 1, 2, 3, ... FOR: $ax^2+bx+c$ the term in partial fraction decomposition is: $\frac{ Ax+B }{ ax^2+bx+c }$ and FINALLY, for: $(ax^2+bx+c)^n$ the partial fration decomposition is: $\frac{ A_1x+B_1 }{ ax^2+bx+c }+\frac{ A_2x+B_2 }{ (ax^2+bx+c)^2 }+...+\frac{ A_nx+B_n }{ (ax^2+bx+c)^n }$ where same as previously, n = 1, 2, 3...

23. zepdrix

Oh boy... abbot gettin all fancy -_- lol

24. abb0t

haters gon hate :P jk :)

25. zepdrix

XD

26. anonymous

haha that helps so much... my prof didn't provide us with this info.

27. abb0t

I think it's because you should have learned this in I think algebra or pre-cal. That's why they usually don't go over partial fraction decomp :P

28. anonymous

:( we had a whole lesson on partial fractions and he seemed to miss "how to split it up"

29. zepdrix

Need help finding any of the unknown constants? Or you tearing through this beast just fine? :O

30. anonymous

haha i had to reboot my comp, but i will slay him now :)

31. anonymous

oh man, how do i solve for the constants?? I got A = -1, but i'm not sure what to use for x to solve for the others

32. anonymous

or should I expand everything first like you said?

33. zepdrix

Yah you have to expand them all out :c sucks.

34. zepdrix

I think.

35. anonymous

haha, well, i will give it a shot.

36. anonymous

too bad wolfram wont do it for me

37. anonymous

I got something ridiculously long

38. zepdrix

I got A=-2 I think... Hmm

39. zepdrix

Oh you square the square thingy, nevermind -1 sounds correct :)

40. anonymous

wait i think it's positive 1 now..

41. anonymous

oh nope.. -1

42. anonymous

haha, second guessing,

43. anonymous

I have no idea how to solve the other constants.

44. zepdrix

\large \begin{align*}4x=&-(x^4+2x^2+1)\\&+(Bx^4+Bx^3+Bx^2+Bx\\&+Cx^3+Cx^2+Cx+C)\\&+(Dx^2+Dx+Ex+E)\end{align*}

45. zepdrix

Did you get something like this? I ran through it a little quickly, so I hope I didn't make any mistakes.

46. anonymous

yep thats what i got.

47. zepdrix

From here, we'll EQUATE LIKE TERMS. So see how we have $$0x^4$$ on the left? we'll set that equal to all the $$x^4$$'s on the right.$\large 0=-x^4+Bx^4$Dividing both sides by x gives us,$\large 0=-1+B$

48. zepdrix

I should have written $$0x^4$$ on the left to stay consistent. Oh well..

49. anonymous

I don't understand what you mean by 0x^4 and where it should be.

50. anonymous

4x is on the left...

51. zepdrix

You equate all terms that of the same degree. There is a 4x on the left. But that is X to the FIRST degree. We'll equate that with all the of FIRST degree X's on the right.

52. anonymous

ohhhhhh

53. zepdrix

You have NO 4th degree x's on the left. So you equate those with the 4th degree x's on the right.

54. anonymous

:) i get ya now.

55. zepdrix

To be honest the hardest part of the problem is over now :) Now we just need to be careful, matching up the orders correctly so we get the right constants.

56. anonymous

ha ok. I've never heard of equating the like terms before..

57. zepdrix

Ya none of my teachers have really explained that step to me very well... So I wouldn't know how to justify it :P

58. anonymous

|dw:1358740342133:dw|

59. anonymous

|dw:1358740410506:dw|

60. zepdrix

That's kinda funny... I don't think we'll even end up using the 4x to find our unknown constants lol

61. anonymous

we dont need it to solve for E?

62. zepdrix

Oh I suppose we could! My bad. I was thinking of matching up the 0th degree X's to find E. But we can match up the 1st degree X's also.

63. zepdrix

Since there's an Ex and also an E term.

64. zepdrix

So either will work.

65. anonymous

oh, 0th degree would be easier... then |dw:1358740648463:dw|

66. zepdrix

looks good c: I came up with the same values.

67. anonymous

to clarify: A = -1 B = 1 C = -1 D = 2 E = 2

68. zepdrix

mhm

69. anonymous

$\large \color{orangered}{\frac{4x}{(x^2+1)^2(x+1)}} = \frac{ -1 }{ (x+1) } + \frac{ x-1 }{ (x^2+1) } + \frac{ 2x+2 }{ (x^2+1)^2 }$

70. zepdrix

So that's the contents of our new integral? k looks good.

71. anonymous

and now i can integrate it right?

72. zepdrix

Yes, we might have to do a little work to integrate some of the pieces. Like, we'll have to write the last 2 terms... umm.. we'll need to split up the fractions.

73. zepdrix

We're going to end up with a bunch of natural logs and arctangents i think.

74. abb0t

So, what did I miss?

75. anonymous

|dw:1358740971154:dw|

76. zepdrix

$\int\limits -\frac{1}{x+1} + \color{green}{\frac{x}{x^2+1}-\frac{1}{x^2+1}} + \color{cornflowerblue}{\frac{2x}{(x^2+1)^2}+\frac{2}{(x^2+1)^2}}dx$

77. zepdrix

Understand what I mean about splitting up the fractions?

78. anonymous

yep, works for me!

79. zepdrix

Yay! So integrate! c: I'm not actually sure what the last term is going to give us. I can't think of it off the top of my head lol

80. abb0t

I think it's arctan?

81. zepdrix

naw, see how the bottom is being squared? It requires a trig sub and then some fussing with D':

82. anonymous

ahh, i'll figure it out from here guys, thanks for the help!!!!

83. anonymous

wish i could give you both medals

84. zepdrix

yay team \c:/