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Homework Help: How to integrate -2x/ (1+x^2)

  1. Feb 3, 2008 #1
    1. The problem statement, all variables and given/known data

    how do i integrate:
    (-2x)/(1+x^2)^2

    2. Relevant equations



    3. The attempt at a solution
    I used the substitution u=1+x^2
    then i got that the integral is
    1/[2(1+x^2)^2 +C

    but i'm not sure if that's correct

    thank you
     
  2. jcsd
  3. Feb 3, 2008 #2

    morphism

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    That substitution is a good idea, but it looks like you're making a mistake afterwards. Maybe if you post your working we'll be able to show you where you went wrong.
     
  4. Feb 3, 2008 #3
    u=1+x^2
    so dx=du/2x
    so =integral(1/2u^2)du
    =integral{1/[2(1+x^2)^2}
    and now im stuck because i realised where i went wrong
    so how do i continue from here?
     
  5. Feb 3, 2008 #4

    HallsofIvy

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    First the 2 in the denominator of dx= du/2x will cancel the 2 in the original integral: you should have "integral (-1/u^2)du.

    (Some people find it easier not to solve for dx: du= 2x dx and you can substitute directly for the "2xdx" in the original integral. And don't forget the "-".)

    More important, why in the world did you go back to the 1+ x2? Don't undo your substitution until after you have integrated!

    Can you integrate [itex]\int (1/u^2) du[/itex]? (Hint: that's a power of u.)
     
  6. Feb 3, 2008 #5
    the integral is 1/u so thats 1/(1+x^2)
    is that the answer?
     
  7. Feb 3, 2008 #6

    Hootenanny

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    Not quite,

    [tex]\frac{d}{du}\left(\frac{1}{u}\right) = -\frac{1}{u^2} \neq \frac{1}{u^2}[/tex]

    Hint,

    [tex]\int\frac{du}{u^2} = \int u^{-2} du[/tex]
     
  8. Feb 3, 2008 #7
    [tex]\int\frac{-2x}{(1+x^2)^2}dx=\int\frac{-2x}{(1+x^2)^2}\frac{d(1+x^2)}{2x}=\int\frac{-1}{(1+x^2)^2}d(1+x^2)=-\frac{(1+x^2)^{-2+1}}{-2+1}+C=\frac{1}{1+x^2}+C[/tex]
    where [tex]d(1+x^2)=2xdx[/tex]
    [tex]dx=\frac{d(1+x^2)}{2x}[/tex]
     
  9. Feb 3, 2008 #8
    no i dont have to integrate u^-2
    i have to integrate -u^-2
    which is 1/u ???
     
  10. Feb 3, 2008 #9
    yes i thought that was right. thank you
     
  11. Feb 3, 2008 #10

    Hootenanny

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    Ahh, my mistake, you are indeed correct. Sorry :redface:
     
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