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Improper Integral

  1. Mar 7, 2005 #1
    Evaluate the integral:
    [tex]\int^{\infty}_{0}\frac{x\arctan{x}}{(1+x^2)^2}dx[/tex]

    can anybody give me some hint? :cry:
    Thanks in advance
     
  2. jcsd
  3. Mar 7, 2005 #2

    Galileo

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    Integrating by parts and substituting [itex]u=\arctan x[/itex] will work.
     
  4. Mar 7, 2005 #3

    dextercioby

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    I hope u know how to choose the "u" and the "dv' for the part integration...

    Daniel.
     
  5. Mar 7, 2005 #4
    I did like this:

    Let [tex]u=\arctan{x} \Rightarrow du=\frac{dx}{1+x^2}[/tex]

    Let [tex]dv=\frac{x}{(1+x^2)^2}dx \Rightarrow v=-\frac{1}{2(1+x^2)}[/tex]

    [tex]\int{udv}=uv-\int{vdu}
    =-\frac{\arctan{x}}{2(1+x^2)}dx+\frac{1}{2}\int{\frac{dx}{(1+x^2)^2}}
    [/tex]

    and then i'm blocked here:
    [tex]\int{\frac{dx}{(1+x^2)^2}[/tex]
    :cry: :cry: :cry:
     
  6. Mar 7, 2005 #5

    dextercioby

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    Use the substituion that Galileo prescribed...Denote the second integral by I:

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

    Make the substitution:

    [tex] x=\tan u [/tex] and say what u get...

    Daniel.
     
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