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Integrals of rational functions

  1. Jan 2, 2008 #1
    i'm trapped with a problem: [tex]\int\frac{dx}{x\sqrt{2-x-x^2}}[/tex].

    i think this problem could be solved by subtitutions: [tex]\ x+\frac{1}{2}=\frac{3}{2}sint[/tex] and [tex]\ u=tan\frac{t}{2}[/tex].
    and finally we would get an expression in [tex]\ u[/tex]: [tex]\frac{\sqrt{2}}{4} log\left|\frac{2\sqrt{2}+u-3}{2\sqrt{2}-u+3}\right|[/tex]
    (am i right so far?)

    however i find it difficult and tedious to write the result in x and get the final answer.

    does anyone know how to evaluate this integral in an alternative way?

    Thanks a lot.
  2. jcsd
  3. Jan 2, 2008 #2
    another question is how to evaluate the integral [tex]\int\frac{\sqrt{2-x-x^2}}{x^2}dx[/tex].
    i used the method of integration by parts, anyone knows some smarter way to do it?
  4. Jan 2, 2008 #3


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    yes this is the correct substitution. my answer is slight different from yours, but it could be just me not doing it carefully, or they are actually the same but written in slightly different form. Anyway method is correct.
  5. Jan 2, 2008 #4
    thanks mjsd.

    i know that the method itself is correct, but it just seems a little tedious, especially when we need to write the final answer in variable x.

    is there any other substitution we could use to attack this integral? or any alternative methods rather than the routine process to integrate rational function?
  6. Jan 2, 2008 #5


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    when I have time I may try to do this again, but I do not believe it is overly complicated when you put it in terms of x. (well, that depends on your definition of "complicated"), you probably just got to know some tricks to simplify it.
  7. Jan 3, 2008 #6


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    By the way, you do understand that these are not "rational functions", as you said in your title, don't you? Was that just a typo for "radical functions"?
  8. Jan 3, 2008 #7

    Gib Z

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    It may be a tiny bit 'tedious' to write again in terms of x, but not difficult, and no where near as tedious as have doing that integral in the first place..

    [tex]x+\frac{1}{2}=\frac{3}{2}\sin t[/tex] so [tex] \sin t = (x+\frac{1}{2})(\frac{2}{3}) = \frac{2x+1}{3}[/tex] hence, [tex]t= \arcsin \left(\frac{2x+1}{3}\right)[/tex]


    [tex]u = \tan \left( \frac{\arcsin \left(\frac{2x+1}{3}\right)}{2} \right)[/tex].
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