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How to integrate this?

  1. May 22, 2006 #1
    [tex]\int{\frac{x^3+x^2+x-1}{x^2+2x+2}dx}[/tex]

    I divided the polynomails and got:

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

    This becomes:

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

    If I've done this right, how do I integrate the third term?

    The arctan should have the (x+2) in brackets but I'm not too skilled in the use of Latex.
     
    Last edited: May 22, 2006
  2. jcsd
  3. May 22, 2006 #2
    You can write the denominator as [itex](x+1)^2 + 1[/itex]. Thats one way out.
     
  4. May 22, 2006 #3
    [tex]\int{(x-1)dx}+\int{\frac{x+1}{x^2+2x+2}dx}[/tex]

    You may consider trying a u substition. It may turn up producing something that looks familiar.
     
  5. May 22, 2006 #4
    [tex]\int{\frac{x+1}{x^2+2x+2}dx}[/tex]

    There is actually a simple way of integrating this expression. Try differentiating the denominator. What do you observe?
     
  6. May 22, 2006 #5

    GCT

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    Just change the denominator to (x+1)^2+1,

    x+1=tan@

    dx=sec^2 @ d@

    You can then integrate

    tan@sec^2@/(tan^2 @ +1) d@ which becomes

    tan@sec^2@/sec^2 @ d@

    thus you're left to integrate tan@ d@

    I may have messed up on some trig identities here, it's been a while since I've worked with them.
     
  7. May 22, 2006 #6

    Hootenanny

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    I would recommend pizzasky's thoughts / process. Think about this integral;

    [tex]\int\frac{f ' (x)}{f(x)} \;dx[/tex]

    ~H

    Edit: Thank you Gokul, unfortunatly that is a mistake I make often, infact I made it today in a mock exam but I'm lucky they don't penalise it in Physics (they did in my maths exam :frown: )
     
    Last edited: May 22, 2006
  8. May 22, 2006 #7

    Gokul43201

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    GCT, it can be done your way, but the observation pointed out by pizzasky makes the computation trivial (you have the answer in a single step).

    Edit : Didn't read Hoot's post when I wrote this.
     
    Last edited: May 23, 2006
  9. May 22, 2006 #8

    GCT

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    silly me, failed to catch on to that.
     
  10. May 22, 2006 #9
    Ah, I see. Separating the second integral isn't needed because I can do it with substitution. Thank you.
     
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