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Partial fractions with fractional powers

  1. Nov 26, 2008 #1
    1. The problem statement, all variables and given/known data

    How does one integrate e.g. [itex]\frac{1+x}{(2+x)^{3/2}}[/itex] by partial fractions?

    3. The attempt at a solution

    I have no idea about this. I've never seen this technique applied with fractional powers before.
     
  2. jcsd
  3. Nov 26, 2008 #2
    This looks interesting, I have never done it but my guess would be to make a substitution.
     
  4. Nov 26, 2008 #3

    gabbagabbahey

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    Hint: [itex]x+1=(x+2)-1[/itex]:wink:
     
  5. Nov 26, 2008 #4
    Nice catch gabba, cant believe I didnt see that ^_^. Though, does that satisfy it being a partial fraction after you simplify/reduce? (Looks up the def.)
     
  6. Nov 26, 2008 #5
    Yes, nice one. Is splitting the numerator in this way a special case of partial fractions? Not that it really matters....
     
  7. Nov 27, 2008 #6

    HallsofIvy

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    No. That is NOT using "partial fractions". This can be integrated but not by partial fractions.
     
  8. Nov 27, 2008 #7
    is the ans
    2 [tex]\sqrt{x+2}[/tex] (1+[tex]\frac{1}{x+2}[/tex])
     
    Last edited: Nov 27, 2008
  9. Nov 27, 2008 #8

    gabbagabbahey

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    Yup :smile:
     
  10. Nov 27, 2008 #9

    gabbagabbahey

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    Sure it is: if you let [itex]u=(2+x)^{1/2}[/itex] then both the numerator and denominator are polynomials in powers of [itex]u[/itex]. You can the decompose it into partial fractions by letting [tex]\frac{1+x}{(2+x)^{3/2}}=\frac{u^2-1}{u^3}=A+\frac{B}{u}+\frac{C}{u^2}+\frac{D}{u^3}[/tex] and determining A,B,C and D....or you can simply recognize that the fraction decomposes into [tex]\frac{1}{(2+x)^{1/2}}-\frac{1}{(2+x)^{3/2}}[/tex]

    Either way, it sure seems like "partial fraction decomposition" to me.
     
  11. Nov 27, 2008 #10
    hahas. thx loads for your hint :smile:
    CHEERS!
     
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