## baldymcgee6 Group Title Integral by partial fraction? one year ago one year ago

1. baldymcgee6 Group Title

|dw:1358736744745:dw|

2. abb0t Group Title

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

3. baldymcgee6 Group Title

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

4. baldymcgee6 Group Title

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

5. baldymcgee6 Group Title

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

6. abb0t Group Title

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

7. baldymcgee6 Group Title

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

8. baldymcgee6 Group Title

how do we solve for all those?

9. abb0t Group Title

You get rid of the denominators.

10. abb0t Group Title

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

11. zepdrix Group Title

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

12. baldymcgee6 Group Title

ahh, thank you zepdrix, was a little confused

13. baldymcgee6 Group Title

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

14. abb0t Group Title

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

15. baldymcgee6 Group Title

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

16. baldymcgee6 Group Title

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

17. zepdrix Group Title

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 Group Title

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

19. zepdrix Group Title

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. baldymcgee6 Group Title

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

21. zepdrix Group Title

Oh i see c:

22. abb0t Group Title

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 Group Title

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

24. abb0t Group Title

haters gon hate :P jk :)

25. zepdrix Group Title

XD

26. baldymcgee6 Group Title

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

27. abb0t Group Title

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. baldymcgee6 Group Title

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

29. zepdrix Group Title

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

30. baldymcgee6 Group Title

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

31. baldymcgee6 Group Title

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. baldymcgee6 Group Title

or should I expand everything first like you said?

33. zepdrix Group Title

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

34. zepdrix Group Title

I think.

35. baldymcgee6 Group Title

haha, well, i will give it a shot.

36. baldymcgee6 Group Title

too bad wolfram wont do it for me

37. baldymcgee6 Group Title

I got something ridiculously long

38. zepdrix Group Title

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

39. zepdrix Group Title

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

40. baldymcgee6 Group Title

wait i think it's positive 1 now..

41. baldymcgee6 Group Title

oh nope.. -1

42. baldymcgee6 Group Title

haha, second guessing,

43. baldymcgee6 Group Title

I have no idea how to solve the other constants.

44. zepdrix Group Title

\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 Group Title

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

46. baldymcgee6 Group Title

yep thats what i got.

47. zepdrix Group Title

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 Group Title

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

49. baldymcgee6 Group Title

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

50. baldymcgee6 Group Title

4x is on the left...

51. zepdrix Group Title

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. baldymcgee6 Group Title

ohhhhhh

53. zepdrix Group Title

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

54. baldymcgee6 Group Title

:) i get ya now.

55. zepdrix Group Title

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. baldymcgee6 Group Title

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

57. zepdrix Group Title

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. baldymcgee6 Group Title

|dw:1358740342133:dw|

59. baldymcgee6 Group Title

|dw:1358740410506:dw|

60. zepdrix Group Title

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

61. baldymcgee6 Group Title

we dont need it to solve for E?

62. zepdrix Group Title

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 Group Title

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

64. zepdrix Group Title

So either will work.

65. baldymcgee6 Group Title

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

66. zepdrix Group Title

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

67. baldymcgee6 Group Title

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

68. zepdrix Group Title

mhm

69. baldymcgee6 Group Title

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

70. zepdrix Group Title

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

71. baldymcgee6 Group Title

and now i can integrate it right?

72. zepdrix Group Title

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 Group Title

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

74. abb0t Group Title

So, what did I miss?

75. baldymcgee6 Group Title

|dw:1358740971154:dw|

76. zepdrix Group Title

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

77. zepdrix Group Title

Understand what I mean about splitting up the fractions?

78. baldymcgee6 Group Title

yep, works for me!

79. zepdrix Group Title

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 Group Title

I think it's arctan?

81. zepdrix Group Title

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

82. baldymcgee6 Group Title

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

83. baldymcgee6 Group Title

wish i could give you both medals

84. zepdrix Group Title

yay team \c:/