1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Ahh, my mistake, you are indeed correct. Sorry :redface:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: How to integrate -2x/ (1+x^2)
Loading...