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Tough Integration Problem Help

  1. Aug 13, 2015 #1
    < Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

    1) ∫ dx/((x^(2))(x^(2)+4)^(1/2))


    2) I'm really stumped on this integral. I've tried several different methods of integration, but I kept getting stuck.


    3) The problem doesn't look like it needs trig substitution, and neither integration by parts, nor substitution seem to work, so I tried rationalizing the denominator to make it easier to decompose the fraction...

    ∫ ((x^(2)+4)^(1/2))/((x^(2))(x^(2)+4)) dx

    then to decompose the fraction, I set it up as...

    (Ax+B)/(x^(2)) + (Cx+D)/(x^(2)+4) = ((x^(2)+4)^(1/2))/((x^(2))(x^(2)+4))
    (Ax+B)(x^(2)+4) + (Cx+D)(x^(2)) = ((x^(2)+4)^(1/2))

    From here, however, I cant figure out the numerators of the two fractions. This is where I get stuck. Any help is greatly appreciated, thanks!
     
    Last edited by a moderator: Aug 13, 2015
  2. jcsd
  3. Aug 13, 2015 #2

    BvU

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    Hello Ryan, welcome to PF :smile: !

    What about substituting ## y = 1/x ## ?
     
  4. Aug 13, 2015 #3
    That doesn't seem to cleanly simplify out either unfortunately - at least when I do it. Great idea though, thank you!!
     
  5. Aug 13, 2015 #4

    BvU

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    Well, what DO you get ? Show your work and perhaps we can find a way out.
     
  6. Aug 13, 2015 #5

    BvU

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    By the way, because of the brackets in post # 1 you still have me wavering a bit between $$\int {dx\over x^2 \sqrt{x^2+4}} \quad {\rm and } \quad \int {\sqrt{x^2+4}\over x^2 }dx $$but I suppose you mean the first of these (because of the ∫ ((x^(2)+4)^(1/2))/((x^(2))(x^(2)+4)) dx)

    By the way: do you see how much clearer a little typesetting makes these expressions ! Well worth learning a bit of TeX !
     
  7. Aug 13, 2015 #6

    SteamKing

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    And you would be wrong.

    There is a trig substitution which works, but you'll need a trig identity to get you started.

    ##sec^2(x) = 1 + tan^2(x)##
     
  8. Aug 13, 2015 #7
    Take x2 out from the root . Now put 1 / x2 as say u . And solve .
     
  9. Aug 15, 2015 #8
    I also think the OP means ##\int {dx\over x^2 \sqrt{x^2+4}}##.

    If you want to do it without StreamKing's trig substitution (which is the easiest, most direct way), you can rearrange the integrand a little. Note that

    $$ \frac{4}{x^2 \sqrt{x^2+4}} \quad = \quad \frac{x^2+4}{x^2 \sqrt{x^2+4}} \quad - \quad \frac{1}{\sqrt{x^2+4}} $$

    The way I found this "clever" method is by reverse engineering it. I did the obvious trig substitution, ## x = 2\tan z ##, got the answer, and then differentiated it to see how the answer must be rearranged to get ##\frac{1}{x^2 \sqrt{x^2+4}}##.
     
  10. Aug 15, 2015 #9
    I respectfully disagree . It might strike the easiest , but I don't think that method is either the shortest , or the easiest .
     
  11. Aug 15, 2015 #10

    Ray Vickson

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    Just decompose 1/(x^2 * (x^2 + 4)) and then multiply by sqrt(x^2 + 4) later! If you write y = x^2 you just have to decompose 1/(y*(y+4)), and that is easy. Then put back x^2 in place of y.
     
  12. Aug 15, 2015 #11
    You're right, it's definitely not the shortest. Your way is. That is very nice. I just meant that trig substitutions pretty much always work and don't require much thought. So you can rely on them when more elegant methods don't jump out at you. Once you have already seen the "correct" way to do it, of course it becomes the easiest.
     
  13. Aug 16, 2015 #12

    verty

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    Is that legal? I'm thinking of bounds like ##\int_{-\pi}^{\pi}##. If you make that substitution, the bounds become equal.

    Something different now, this question demonstrates that it is not worthwhile making a hyperbolic substitution (I have not seen a case where it is ever worthwhile). This looks like a natural example where a hyperbolic substitution, if it was ever worthwhile, would work well. But try it and you'll quickly see that it is much worse than the usual trig sub.
     
    Last edited: Aug 16, 2015
  14. Aug 16, 2015 #13
    If it was a definite integral , I would be breaking the integral from - π to 0 , and from 0 to π ( Why ? ) .

    Does this answer your question ?
     
  15. Aug 16, 2015 #14

    verty

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    Actually you would need something like ##\int_{-\infty}^{-\pi} + \int_{\pi}^{\infty}##. It doesn't seem worth the hassle to me. And I wonder if it would be literally false to say ##\int ... dx = \int ... du + C##...

    Edit: Hmm, I definitely got this wrong. I need to think more about this.
     
    Last edited: Aug 16, 2015
  16. Aug 17, 2015 #15
    Your edit - I'm not sure what you are referring to , so I'll just answer the part in this post .

    First part - Limits would be from -pi to 0 and 0 to +pi .

    Other part - Why so ? Cannot two different graphs have the same area when the limits of the two functions are different ?
     
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