## anonymous one year ago Express following in partial fractions...evaluate the given integral

1. anonymous

|dw:1438068450403:dw|

2. anonymous

@UsukiDoll

3. UsukiDoll

oh crap... let me expand (x-1)^2(x+1) first with wolfram and see if it comes back as factorable all the way.

4. UsukiDoll

ok it's all factorable so we have single letters to deal with.. one problem whenever we have a repeater... we have to take the original and build up the exponents.

5. UsukiDoll

for example |dw:1438068733033:dw| then take lcd of all of these guys

6. UsukiDoll

luckily the problem we are given isn't harsh like that (though I did s^5 at one point)

7. UsukiDoll

8. anonymous

ok hold on

9. anonymous

$\frac{ 2x }{ (x-2)^2(x+1) }=\frac{ A }{ x+1 }+\frac{ B }{(x-2)^2 }$ simplify $2x=\frac{ A(x+2 )((x-2)^2) }{ x+2 } + \frac{ B((x-2)^2(x+2)) }{ (x-2)^2 }$

10. anonymous

and then

11. UsukiDoll

STOP!

12. UsukiDoll

the A part is fine.. the one with the exponent.. no...

13. anonymous

$2x=A(x-2)^2+B(x+1)$ x=2 gives $4=A(0)^2+B(3)$ B=4/3

14. anonymous

ohh

15. UsukiDoll

whenever you have the exponent .. you have to write the single version first and then the exponent (depending on the number) look at this example |dw:1438069380011:dw|

16. UsukiDoll

I wrote the single version s and then work the exponent up s s^2 s^3 s^4 then you got to take lcd of all those fractions

17. UsukiDoll

|dw:1438069450631:dw|

18. UsukiDoll

I think order matters too... |dw:1438069483088:dw|

19. UsukiDoll

the lcd for this is $(x+1)(x-1)^2$ alright take the lcd ;)

20. UsukiDoll

ahem... take the lcd for |dw:1438069618368:dw|

21. anonymous

alright alright, give me a sec:)

22. anonymous

|dw:1438069792243:dw|

23. UsukiDoll

stop setting 2x = that's not going to work

24. UsukiDoll

the lcd part is good though

25. UsukiDoll

|dw:1438069902990:dw|

26. UsukiDoll

k expand all of that

27. UsukiDoll

*eats popcorn while I wait*

28. anonymous

I dont get this at all....whats wrong with me!!

29. UsukiDoll

oyyyyyyyy........

30. UsukiDoll

|dw:1438070102575:dw|

31. UsukiDoll

32. anonymous

whats does foil mean':)?

33. UsukiDoll

foil - first outer inner last... okkk expand :)

34. anonymous

Maybe I shouldve told u havent slept in 2 days...

35. UsukiDoll

then you should take a break from math and sleep

36. anonymous

its 10am...my exam is in 1 month from today....I have alot to do...due to circumstances...I cant sleep til 12hours from now,

37. UsukiDoll

try finishing the problem?

38. anonymous

so expanding c gives x^2-2x+1

39. anonymous

for A gives x^2-1

40. UsukiDoll

|dw:1438070573305:dw| |dw:1438070618427:dw|

41. UsukiDoll

yeah...

42. UsukiDoll

|dw:1438070677726:dw|

43. UsukiDoll

now distribute the letters

44. anonymous

Im really sry, can you finish it? I cant think properly, after this im going to eat and take a break and then start this again fresh.

45. UsukiDoll

the problem is that I can't give direct answers... so can't finish it without getting busted.

46. anonymous

@aric200

47. UsukiDoll

@Xlegalize can you come back over here so we can finish this please?

48. anonymous

im here

49. UsukiDoll

ok so the last step we left off is |dw:1438073386251:dw| so what we need to do next is to distribute the letters

50. UsukiDoll

I'll give an example $A(x^2-1) = Ax^2-A$

51. UsukiDoll

then similarly for B and C $B(x+1) \rightarrow Bx+B$ $C(x^2-2x+1) \rightarrow Cx^2-C2x+C$

52. UsukiDoll

yeahhh that 2xC combination looks gross... and awkward XD anyway now we can put that long string in the numerator .

53. UsukiDoll

$\large \frac{Ax^2-A+Bx+B+Cx^2-2xC+C}{(x+1)(x-1)^2}$

54. UsukiDoll

|dw:1438073620766:dw| on the original equation, we have a 2x so that would mean that all equations with x would be equal to 2. whereas the plain letters and the letters with x^2 will be set equal to 0

55. UsukiDoll

$-A+B+C =0$ $B-2C=2$ $A+C =0$

56. UsukiDoll

so now we have these systems of equations and we need to solve for A,B,C once we get A B C we can plug it back into the equation from earlier

57. anonymous

B=2+2c A=2+2c+c A+c=0 2+2c+c+c=0

58. UsukiDoll

let's try adding the first and third equation up $-A+B+C =0$ $A+C =0$ the A cancels out and we are left with B+2C= 0

59. anonymous

ok that is easier.

60. UsukiDoll

so now adding B-2C=2 B+2C=0 we have 2B=2 B = 1

61. UsukiDoll

so now that B = 1 the only place I can put that is the second equation if I put this in the first equation I'm stuck.. because I don't know what C or A is...yet

62. UsukiDoll

B-2C=2 1-2C=2 |dw:1438074079535:dw|

63. UsukiDoll

so B = 1, C = -1/2 now we can get our A picking out an easier equation with A is the best method to getting what A is

64. anonymous

a=1/2

65. UsukiDoll

|dw:1438074167001:dw|

66. UsukiDoll

that's right.. now that A = 1/2, B = 1, and C = -1/2 we need to check if they satisfy the three equations. If they do then we can go back, plug and chug into the earlier equation, and use integration

67. UsukiDoll

$A+C =0$ since A = 1/2 and C = -1/2 $\frac{1}{2}-\frac{1}{2}=0$ $0=0$ $-A+B+C =0$ $-\frac{1}{2}+1-\frac{1}{2} =0$ $-\frac{1}{2}-\frac{1}{2}+1 =0$ $-\frac{1}{2}-\frac{1}{2}+1 =0$ $-\frac{2}{2}+1 =0$ $-1+1 =0$ $0 =0$ $B-2C=2$ B = 1 $1-2\frac{-1}{2}=2$ $1+\frac{2}{2}=2$ $1+1=2$ $2=2$ ok after that long verification latex string we can plug our values into the earlier equation and integrate

68. UsukiDoll

|dw:1438074628205:dw|

69. UsukiDoll

so after all that lcd solving for a b c we can finally integrate... YAY! 2 of them are logs.. one is u substitution

70. UsukiDoll

|dw:1438074757028:dw| so the red portion is the $\frac{1}{(x-1)^2}$ part. we have to do a bit more work on it

71. anonymous

Go on:)

72. anonymous

yeah I also got those parts.

73. anonymous

the last one takes more work.

74. UsukiDoll

yeah x.x now.. using u sub. u = x-1 du = 1 dx

75. UsukiDoll

|dw:1438074892358:dw| I wanna recheck that just to be sure

76. anonymous

yep thats correct.

77. UsukiDoll

thanks :) so final answer after all that jazz is |dw:1438075034959:dw|

78. anonymous

I will take a break and go through this all again. So Im 100 on this THANKS!!!!

79. UsukiDoll

you're welcome :)