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Problem on integration

  1. Feb 28, 2005 #1
    can anyone help me with this,

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

    thanks.

    I've tried with integration by parts but does not work..
     
  2. jcsd
  3. Mar 1, 2005 #2
    Hmm... I don't quite remember how to use partial fractions for that... but I remember there was a way (at least with a similar problem). You could try

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

    and use partial fractions from there (but it gets a bit "hairy," as my math teacher always says).
     
  4. Mar 1, 2005 #3
    Did you try partial fractions?

    edit: nevermind read above :)
     
  5. Mar 1, 2005 #4
    Partial fractions work only for proper fractions. This is not a proper fraction in the sense that the degree of the numerator is the same as the degree of the denominator. So to fix this little problem you need to use long division after which you get.

    [tex] 1-\frac{1}{x^2+1} [/tex]

    Which is now easy to integrate.

    Regards
     
    Last edited: Mar 1, 2005
  6. Mar 1, 2005 #5
    Use x^2/(1 + x^2) = (x^2 + 1 - 1)/(1 + x^2) = 1 - 1/(1 + x^2) and integrate that instead...
     
  7. Mar 1, 2005 #6

    dextercioby

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    In this case partial fractions would lead indeed to a decomposition containing complex numbers.Therefore the method is not good.

    Daniel.
     
  8. Mar 1, 2005 #7
    I did this on my calculator and got a simple answer. I think you can do integration by parts. What have you tried?
     
  9. Mar 1, 2005 #8
    Integration by parts isn't necessary. Townsend clearly explains the best way to do this.

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

    Hint for the second part: trig inverses.
     
  10. Mar 1, 2005 #9
    thank you everyone for the help, i understand now :smile:
     
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