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Need help, Integration of irrationals.

  1. Jun 26, 2007 #1
    Hi, i am trying so hard to find a way out of integrating this irrational function, but i cant jus figure out how to do it.
    Here it is:
    integ dx/(x^2 +x+1)(X^2 +x-1)^(1/2),
    well here is what i tried. After some calculations i transformed the denominator to this form

    integ dx/[(x+1/2)^2 +3/4][(x+1/2)^2 -5/4]^1/2, then i substituted x+1/2=t
    where dx=dt, so i got

    integ dt/(t^2 +3/4)(t^2 -5/4)^1/2 ,, then i tried a lot more but i cant just come to a solution, it keeps expanding instead of coming up to something. I think that from here i am not taking the right path.
    So, can you guys give me some hints on how to go about integrating this??


    P.S. The answer is in the textbook, i do not know whether it is correct or not, and it is:

    I= 1/6^(1/2) ln[(3(x^2 +x-1))^(1/2)+(2x+1) 2^(1/2)]/[(3(x^2 +x-1))^(1/2)-(2x+1) 2^(1/2)]
    Last edited: Jun 26, 2007
  2. jcsd
  3. Jun 26, 2007 #2
    This is unreadable to a casual reader. If anyone helps you, be sure to thank them for taking their time to figure everything out.

    If you want more help, please learn latex formating. :)
  4. Jun 26, 2007 #3


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    I think your integral is
    [tex]\int \frac{dx}{(x^2+ x+ 1)\sqrt{x^2+ x- 1}}[/tex]

    A standard technique would be to complete the square in the radical and then look for a trig substitution.
  5. Jun 26, 2007 #4
    Yeah, i know. I really appologize for writing this way, but i don't know yet how to use latex. I guess i'll have to learn soon.
  6. Jun 26, 2007 #5
    Yeah, i have completed the square in the radical, but i did not think of taking a trig supstitution. I'll try that tomorrow, couse it's too late now. Thnx for now, if i can't yet figure it out tomorrow, than surely i'll be back.
  7. Jun 27, 2007 #6
    Well, i tried a trig substituton, i took this substitution

    t= [(5/4)^(1/2)] / sin u,

    However the asnwer that i got is different from the one in the book, and also different from the one that "The integrator" ,an online version of mathematica as you may know, gives. But i think the idea works, i dont think i have made any mistakes in the process though, although it is quite likely.
  8. Jun 27, 2007 #7


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    [tex]x^2+ x- 1= x^2+ x+ \frac{1}{4}-\frac{1}{4}-1= x^2+ x+ \frac{1}{4}-\frac{5}{4}= (x+ \frac{1}{2})^2- \frac{5}{4}[/tex]
    A sin substitution would work for something of the form a2- usup]2[/sup] but this is of the form x2- usup]2[/sup]. (I just noticed that you have "/sin u" so that's really a csc substitution.)

    Remembering that sin2x+ cos2x= 1, We can divide both sides by sin2x and get cot2x+ 1= csc2x or cot2x= csc2x- 1 which suggests the substitution [itex]x+1= \sqrt{5}/2 csc(\theta)[/itex]. Then [itex]dx= -\sqrt{5}/2 csc(\theta)cot(\theta) d\theta[/itex]. Of course, [itex]\sqrt{(x+1)^2- 5/4}= \sqrt{5}/2 cot(\theta). The hard part is the x2+ x+ 1 but fortunately the "x2+ x" is the same so we can complete the square in the same way: x2+x+1= (x+1)2+ 3/4. That polynomial becomes [itex](5/2)csc^2(\theta)[/itex]+ 3/4[/itex].
  9. Jun 27, 2007 #8
    Well, this is almost exactly what i did, with the exception that i did not write the 1/sinu thing as cscu.
  10. Jun 28, 2007 #9


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    Ok, can you post your final answer here, so that we may verify it for you?

    Btw, you should note that there are infinite number of anti-derivatives to f(x), all of which differ each other by a constant. E.g, -cos2(x), sin2x, and (-1/2)cos(2x) are all anti-derivatives of sin(2x), but they differ from each by a constant.
  11. Jun 28, 2007 #10
    Yeah, here it is, but i dont know latex formating, so it is going to be a little messy.

    2/(6^1/2) log[2(2^1/2) -(3^1/2)arcsin(5^1/2)/(2x+1)]/[2(2^1/2) +(3^1/2)arcsin(5^1/2)/(2x+1)]
    I dont really know whether this is right or not, it looks like it is but...
    thnx in return
  12. Jun 28, 2007 #11

    Gib Z

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    I'm not going to bother, but go to www.calc101.com and put that into the derivative finder. If its the original function, your home free. But you missed the whole point of VietDao29's post, you forgot the Constant :P
  13. Jun 28, 2007 #12
    NO, i got what VietDao29 meant about constant. I do understant that if F(x), and Q(x) are both primitive functions of f(x), than F(x)-Q(x)=C,
    However in my case, i could not show the equivalence of the two solutions, the one that i got and the other which is on the textbook. Thnx for the link, i will try tha later, couse right now i am having some problems inputing it. But i got the idea about how to tackle these kinds of integrals. THNX.
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