## anonymous 3 years ago Hi! Integrate 2x/(x^8+1) dx

1. anonymous

I'll be honest, I can solve that......the solution is not neat......so don't expect anyone here to be able to finish it....it's long...........try wolfram....... but you still can try here though

2. anonymous

can you tell me how to at least start it?

3. anonymous

I feel like this will end up being an ln function, but I think there's a lot more to it than just that I'm sure. Let Ralph help you.

4. anonymous

does anyone know of a completed example of a problem like this I can go online and review or anything?

5. anonymous

|dw:1368859626389:dw|

6. anonymous

That's how you start.

7. anonymous

can you tell me what the J is for? I have never ever seen that used before

8. anonymous

|dw:1368859717310:dw|

9. anonymous

J is just a multiplier. In this case, J = 1 , so you can remove it

10. anonymous

(x^8+1) means: X^9

11. anonymous

I highly doubt that it would be that simple felavin.

12. anonymous

WHY? SMOKEYDABEAR.

13. anonymous

could the next step be to let u=tan(theta) then du=sec^2(theta) d(theta)

14. anonymous

@SinginDaCalc2Blues same thing.........you're looping

15. anonymous

Yeah all that would do is put you right back to where you started

16. anonymous

^^^ that girl is just too smart..........

17. anonymous

Too stupid*

18. anonymous

I'm just here to learn how to do it myself

19. anonymous

are you people calling me stupid? seriously?

20. anonymous

No, I was calling myself stupid xD

21. anonymous

no, I didn't mean that.

22. anonymous

okay, let's figure out an easier way to solve this problem.

23. anonymous

@SinginDaCalc2Blues can you handle recursive integrals?

24. anonymous

okay...well I appreciate the help from everyone but it's hard to understand the context of non-verbal communication sometimes...I was hoping you guys weren't calling me stupid...anyway...

25. anonymous

@SinginDaCalc2Blues we really would never do that.

26. anonymous

i have never heard the term recursive integral...while I probably know what your talking about, I don't know it in those terms...can you describe?

27. anonymous

wait............a minute...why not just polar?

28. anonymous

Polar coordinates? Actually that makes some sense...

29. anonymous

are you saying turn the integral into a polar expression or something?

30. anonymous

@smokeydabear tricky thing is, it returns the integrand to the form we had before....because using cos, cos = 1/sec.....

31. anonymous

my professor is devious!

32. anonymous

Your professor is pretty evil for giving you something like this

33. anonymous

would u=x^2 du=2x dx indefinite integral of 1/(u^4+1) help? (sorry, my latex isn't working)

34. anonymous

@SinginDaCalc2Blues I reduced it to......can you solve this?|dw:1368860855998:dw|

35. anonymous

the int, called complex in

36. anonymous

uhm if k is a constant, what is J again?

37. anonymous

J is the variable of integration, just like U

38. anonymous

U in U substitution

39. anonymous

I just looked it up on wolfram... what is this monstrosity...

40. anonymous

@smokeydabear I told you...........

41. anonymous

How do you know where to even start with something like this? That's always been my problem

42. anonymous

a good place to start is with Vodka :o)

43. anonymous

@SinginDaCalc2Blues I asked you if you knew recursives because as you can see here, the trigs are repeating..which means recursion

44. anonymous

45. anonymous

in 30 minutes i will no longer be able to do integration but I will be able to slurrrr Russian! :o)~

46. anonymous

vodka inhibits your ability to do math.

47. anonymous

despair only temporarily inhibits your ability to do everything

48. anonymous

@smokeydabear recursion works to start. for example|dw:1368861204284:dw|

49. anonymous

Okay, so how would we use that to help us? I'm still confused :s

50. anonymous

crazy problem o:

51. anonymous

well long only :P

52. anonymous

@smokeydabear it saves time because once you know that it will repeat, you can generalize..........

53. anonymous

hey guys...i have an idea...gimme a sec...

54. anonymous

.........meaning............|dw:1368861457508:dw|

55. stamp

u = x^2, du = 2x dx$\int \frac{du}{u^4+1}$

56. anonymous

so, because they occur in trig functions, we can use it for the tan theta

57. anonymous

@stamp did that

58. anonymous

Ohhhhhh I see what you did with the sin's and the pi's @primeralph very clever! =D

59. anonymous

now we just need to learn to use it effectively...

60. anonymous

|dw:1368861979382:dw|

61. anonymous

I gotcha Ralph, I noticed it too but I didn't want to be mean =P

62. anonymous

63. anonymous

Does this little bit of trickyness open any doors?

64. anonymous

yeah what stamp wrote I believe is wrong...it wouldn't be x^4...it would be (x^4)^2

65. anonymous

Well what stamp wrote isn't technically wrong, but it isn't the method to solving this.

66. anonymous

I basically factored out of the denominator an x^2...then did u-sub

67. anonymous

68. anonymous

hi, the integrate is simple: Who can solve this integrate:$\int\limits_{}^{} \frac{\tan Z }{ Z- \frac{ P }{ 4}}$ GAME IS OVER ..........

69. anonymous

hello

70. anonymous

hi...i'm only in calc 2...i don't know what Cauchy-Reimann equations are... I did however get my integral down to 1/(u^3+1)

71. anonymous

In case you love twisting with partial fractions: $\frac{ 2x }{ x^8 +1}=\frac{ 2x }{( x^4-\sqrt{2}x^2+1)( x^4+\sqrt{2}x^2+1)}$

72. anonymous

hi drawer...i think your way is looking promising...i think i was headed toward that...can you look at my .bmp and tell me how to start partial fraction with what i ended with?

73. anonymous

74. anonymous

Then $u^3+1=(u+1)(u^2-u+1)$, use partial fractions.

75. anonymous

what do i do with the numerator?

76. anonymous

$\frac{ 1 }{ u^3+1 }=\frac{ A }{ u+1 }+\frac{ Bu+C }{ u^2-u+1 }$ =>$A(u^2-u+1)+(Bu+C)(u+1)=1$ Then equate the coefficients and you'll get a system of linear equations in terms of A,B, and C. Solve it!

77. anonymous

so then...after I do that, will i have technically several different integrals which will be easier to integrate?

78. anonymous

Yes, you'll see that the degree of the polynomial of the numerator is always 1 less than that of the denominator, which must have sth to do with the natural logarithm (ln) function.

79. shubhamsrg

i havent read much what is written above, its too long, has the question been solved ?

80. anonymous

ok drawar...the answer definitely has some nasty ln's...i will keep plugging away at it...one more question...

81. anonymous

once i figure out what A, B, AND C equate to, don't I just plug those into the numerators and then treat one as a separate integral to solve?

82. shubhamsrg

has anyone tried x^2 =t ?

83. anonymous

yeah shub...wayyy up at the top...we are past that now...but thanks

84. anonymous

Yes @SinginDaCalc2Blues

85. shubhamsrg

well, why you didnt go forward with that method, after that substitution, we'll have dt/(t^4 +1) note that t^4 + 1 = (t^2 + i) (t^2 -i) also note that (t^2 +i) - ( t^ -i) = 2i , a constant

86. anonymous

ok...i will keep plugging away and see if I can get it...this problem has been so crazy! :o)

87. anonymous

are you using imaginary numbers shub?

88. anonymous

is it a definite/indefinite integral in your original question?

89. shubhamsrg

yes, the very same. i can be treated as just another constant

90. anonymous

indefinite

91. anonymous

i have no idea how to use imaginary numbers...i'm only in calc 2..we haven't used those

92. shubhamsrg

I see, well use of iota will lead you to the solution withing 2 more steps, and in the end you can always replace the complex part with hyperbolic functions, or natural logarithms . But anyways, since you havent learnt about that, go with @drawar 's method (:

93. anonymous

I wish I had learned about them...i just hope i can solve this

94. anonymous

$\int\frac{2x}{x^8+1}dx$$=\int \frac{1}{x^8+1}d(x^2)$$=\int \frac{1}{y^4+1}dy$$=\frac{1}{2}\int \frac{y^2+1-y^2 +1}{y^4+1} dy$$=\frac{1}{2}(\int \frac{y^2+1}{y^4+1}dy-\int\frac{y^2 -1}{y^4+1} dy)$$=\frac{1}{2}(\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy-\int\frac{1 -\frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy)$ Consider $\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy$ Let u = $$y-\frac{1}{y}$$ $$du =(1 + \frac{1}{y^2}) dy$$ and $$y^2 + 1/y^2 = (y-\frac{1}{y})^2+2=u^2+2$$ So, $\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy$$=\int \frac{1}{u^2+2}du$ Similarly, for$$\int \frac{1-\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy$$ Let $$v = y+ \frac{1}{y}$$ $dv = 1 - \frac{1}{y^2}dy$$y^2 + 1/y^2 = (y+\frac{1}{y})^2-2=v^2-2$ So, $\int \frac{1-\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy$$=\int\frac{1}{v^2-2}dv$ As a result, $\frac{1}{2}(\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy-\int\frac{1 -\frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy)$$=\frac{1}{2}(\int\frac{1}{u^2+2}du-\int\frac{1}{v^2-2}dv)$And You can integrate it then.

95. anonymous

@roly Thank you for taking the time to illustrate it.