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Difficult integration by parts

  1. Aug 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Calculate the following integral with partial integration:

    [tex]\int\frac{1 - x}{(x^2 + 2x + 3)^2}dx[/tex]

    3. The attempt at a solution

    I guess you need to write the integral in easier chunks but I still fail every time.
     
    Last edited by a moderator: Aug 3, 2013
  2. jcsd
  3. Aug 3, 2013 #2
    Can you write this as partial fractions? or make use of the fact that the bottom can be factorised?
     
  4. Aug 3, 2013 #3

    verty

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    I haven't tried this but letting u = 1/(x^2 + 2x + 3) may get you somewhere.

    Sorry, I've checked this now and it doesn't work. Another angle of attack, probably not the best method though, is u = x + b/2a = x + 1.
     
    Last edited: Aug 3, 2013
  5. Aug 3, 2013 #4

    verty

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    Making the substitution I suggested (u = x + 1), one gets this:

    ##\int{\frac{2-u}{(2+u^2)^2}} du##

    Then, one has a choice of substitutions (tan, cot, sinh). By inspection, sinh won't work. We then have denominator cosh^3, not nice. That leaves tan and cot although it seems tan is never a worse substitution than cot. I hold out hope but... use tan and...

    Best of luck.

    PS. Given the nature of this question, I can offer no further help. A difficult question is meant to be difficult.
     
  6. Aug 3, 2013 #5

    HallsofIvy

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    With a quadratic that cannot be factored (using real numbers) in the denominator, the standard step is to complete the square: [itex]x^3+ 2x+ 3= x^2+ 2x+ 1+ 2= (x+ 1)^2+ 2[/itex] so you can write the integral as [tex]\int\frac{1- x}{((x+1)^2+ 2)^3}dx[/tex].

    Now let t= x+ 1 so that dx= dt and 1- x= 2-(x+1)= 2- t. With that substitution, the integral becomes
    [tex]\int\frac{2- t}{(t^2+ 2)^2}dt= 2\int\frac{dt}{(t^2+ 2)^2}- \frac{t}{(t^2+ 2)^2}dt[/tex]
    For the first integral try [tex]t= \sqrt{2}tan(\theta)[/tex] and for the second, [tex]u= t^2+ 2[/tex].
     
    Last edited by a moderator: Aug 3, 2013
  7. Aug 3, 2013 #6
    Another way to tackle the integral could be to somehow create the derivative of the base for the exponent in the denominator in the numerator. Then, you could use substitution.
     
  8. Aug 3, 2013 #7

    lurflurf

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    ^That is pretty much what they did
    $$\frac{1-x}{(x^2+2x+3)^2}=\frac{2-(x^2+2x+3)^\prime /2}{(x^2+2x+3)^2}$$
    then we need to split it into two parts otherwise we need to use complex numbers
     
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