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A little calculus problem

  1. May 11, 2004 #1
    Hi, ive been trying to solve this integral for a while now, but im still making no headway, any help would be much appreciated, thanks in advance.
    [tex]\int\frac{1}{1-x^3}dx[/tex]
     
  2. jcsd
  3. May 11, 2004 #2

    Zurtex

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    You tried splitting into partial fractions? I am just trying now :smile:
     
  4. May 11, 2004 #3

    Zurtex

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    Not 100% sure on this, you'll have to check it but I think:

    [tex]\frac{1}{1-x^3} = \frac{x + 2}{3x^2+3x+3} + \frac{1}{3-3x}[/tex]

    That looks much easier to integrate if it is true :smile:
     
  5. May 11, 2004 #4
    Couldn't you rewrite it as
    [tex]\int {(1-x)^{-3}} dx[/tex]
    and find the anti-dervative that way?
     
  6. May 11, 2004 #5

    arildno

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    Because it is incorrect.
     
  7. May 11, 2004 #6
    Oh, yeah. I see now. Only the x is cubed, not 1 - x
     
  8. May 11, 2004 #7

    Zurtex

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    From the little I understand from integration (and at a guess you probably understand about the same) I don't think you can integrate that any other way than my suggested:

    [tex]\int\frac{1}{1-x^3}dx = \int \frac{x + 2}{3x^2+3x+3} + \frac{1}{3-3x}dx[/tex]

    It's fairly easy to manipulate that and make it possible to integrate.
     
  9. May 11, 2004 #8
    Im not sure how you got those equalities, I get…

    [tex]\frac{1}{1-x^3} = \frac{x + 2}{-x^4 -2x^2+x+2}[/tex]
     
  10. May 11, 2004 #9

    arildno

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    Corrected:
    [tex]\frac{1}{1-x^3} = \frac{x + 2}{-x^4 -2x^{3}+x+2}[/tex]
    But this is not a partial fractions decomposition!!
     
  11. May 11, 2004 #10
    now im not sure how you are getting
    [tex]+ \frac{1}{3-3x}dx[/tex]
     
  12. May 11, 2004 #11
    im not sure if this is right, but this is what i did...

    [tex]\int\frac{1}{1-x^3}dx[/tex]

    [tex]\int\frac{1}{(1-x)(1+x+x^2)}[/tex]

    [tex]\int((1-x)(1+x+x^2))^{-1}[/tex]

    let u = 1 + x + x^2
    du = 1+2x dx

    [tex]2\int (1+u)^{-1} du[/tex]

    [tex]2(u+\frac{u^2}{2})^{-1}[/tex]

    [tex] 2(1+x+x^2+ \frac{(1+x+x^2)^2}{2})^{-1}[/tex]
     
  13. May 11, 2004 #12
    The only method i could think of for doing this question is using partial fractions.

    With this you obtain:
    A - B = 0
    A + B - C = 0
    A + C = 1
    Therefore,
    C = 2/3
    A = B = 1/3

    and f(x) =
    (1/3) * (1/(1-x) + (x+2)/(1+x+x^2))

    Which you would then integrate. (upon further review, its the same thing as what zurtex got)

    JonF, it cant be done using substitution (or atleast i dont see it). You made an error when you were substituting.
     
    Last edited: May 11, 2004
  14. May 11, 2004 #13
    TI89ed it... didn't look too pretty!
     
  15. May 12, 2004 #14
    Argh, cursed arithmetic
     
  16. May 12, 2004 #15

    Dr Transport

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    [tex] \int{(1-x^{3})}^{-1}dx = -\frac{1}{3}ln(1-x)+\frac{1}{3}ln(x^{2}+2x\cos({\frac{\pi}{3}})+1)
    [/tex]

    It is equation 2.144.2 in Gradshteyn and Ryzhik, pg 64
     
  17. May 12, 2004 #16
    how evil! what an equation... is it any use to... umm... rememorize the table of integrals?
     
  18. May 13, 2004 #17

    HallsofIvy

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    No- you've lost the "-1" power in the first term. That should be

    [tex]\int((1-x)^{-1}(1+x+x^2))^{-1}[/tex]
    and now the substitution doesn't work.

    "how evil! what an equation... is it any use to... umm... rememorize the table of integrals?"

    Believe it or not, it's actually easier to THINK!

    In this case, you use "partial fractions" as you were told to begin with:
    1-x3= (1- x)(1+ x+ x2 so
    [tex]\frac{1}{1-x^3}= \frac{A}{1-x}+ \frac{Bx+C}{1+ x+ x^2}[/tex]
    for some numbers A, B, C.

    Multiply that equation by 1-x3 and you get
    1= A(x2+ x+ 1}+ (Bx+ C)(1-x)

    Since this is true for all x, in particular if we let
    x= 1, 1= 3A so A= 1/3

    x= 0 1= A+ C= 1/3+ C so C= 2/3

    x= -1 1= A+ (-B+C)(-2)= 1/3+ 2B- 4/3 so 2B= 2 or B= 1

    Now we need to integrate
    [tex]\frac{1/3}{1-x}+ \frac{(2/3)x+ 1}{x^2+x+1}[/tex]

    The first of those is just -(1/3)ln|1-x| (that's in Dr Transport's answer)

    To do the second, complete the square: x2+ x+ 1= x2+ x+ 1/4+ 3/4= (x+1/2)2+ 3/4 so we break that into two integrals:
    Now make the linear substitution u= x+ 1/2, du= dx, (2/3)x+ 1= (2/3)u- 1/3 and the integral becomes
    [tex]\frac{(2/3)u}{u^2+ 3/4}- \frac{1/3}{u^2+3/4} [/tex]

    The first of those can be done by the further substitution v= u2+3/4 and the second is an arctangent. That might reduce to what Dr Transport gave but it can be difficult to show that!
     
  19. May 13, 2004 #18

    Zurtex

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    HallsofIvy, you seem to have got your partial fractions wrong. I tried making them into 1 fraction again and it didn't work. I am fairly sure that:

    [tex]\frac{1}{1-x^3} = \frac{(1/3)(x + 2)}{x^2+x+1} + \frac{1/3}{1-x}[/tex]

    As I stated on the first page.
     
  20. May 13, 2004 #19
    The way i was taught to do it was to expand everything out, bring together x^2, x... terms together, factor things out and see what you need. For this example, expanding and combining yields:

    1 = A(x^2 + x + 1) + (Bx + C)(1-x)
    1 = Ax^2 + Ax + A + Bx - Bx^2 + C - Cx
    1 = (A - B)x^2 + (A + B - C)x + (A + C)

    Since you have no x^2 or x terms:
    (A - B) = 0 A = B
    (A + C) = 1 A = 1 - C
    (A + B - C) = 0

    Substitude that all into the last equation:
    2A - C = 0
    2(1 - C) - C= 0
    2 - 3C = 0
    C = 2/3
    A = 1/3
    B = 1/3

    which will give you zurtex's partial fractions.

    you accidently got a negative on that two; otherwise you get the same as above.
     
  21. May 17, 2004 #20
    after looking at this again, if you complete the square on the denominator you obtain:

    (x + 2)/[(x + 1/2)^2 + 3/4]

    using a trig substitution you should be able to solve it
     
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