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Difficult Integration - Apostol Section 6.25 #40

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]\int\frac{\sqrt{2-x-x^2}}{x^2}dx[/itex]

    Hint: multiply the numerator and denominator by [itex]\sqrt{2-x-x^2}[/itex]


    2. Relevant equations
    This is in the Integration using Partial Fractions section, but the last few have not been using Partial Fractions.


    3. The attempt at a solution
    Well, initially I thought that I would just complete the square on the top and then use a substitution such as [itex]x+\frac{1}{2} = \frac{3}{2}\sin u[/itex], but that became pretty complex. Then I took the author's suggestion and multiplied the numerator and denominator by [itex]\sqrt{2-x-x^2}[/itex]. I split the resulting integral into three, two of which were easy and the first which is still very difficult:

    [itex]\int \frac{2}{x^2 \sqrt{2-x-x^2}} dx[/itex]

    The substitution I mentioned earlier still looks most promising, but leads to this:

    [itex]\frac{9}{4} \int \frac{\cos^2 u}{(\frac{3}{2}\sin u + \frac{1}{2})^2} du[/itex]

    This looks like it needs something like [itex]z=\tan \frac{u}{2}[/itex], but that also becomes extremely tortuous. It leads to a degree 6 polynomial on the bottom, and a degree 4 polynomial on the top which can then be solved with partial fractions, but the resulting equations are quite cumbersome.

    Any other suggestions/hints?
     
  2. jcsd
  3. Sep 4, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Try integration by parts:

    [itex]9\int \frac{\cos^2 u}{(3\sin u + 1)^2} du=\int (\frac{3\cos u}{(3\sin u+1)^2}) (3 \cos u) du[/itex]

    ehild
     
  4. Sep 4, 2011 #3
    That does look much better, thanks!
     
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