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Integ prob but I'm stuck

  1. Jun 14, 2007 #1
    I'm trying to do the following:

    int (3x^3 + 4x)/(x^2+1)^2 dx

    I let u = x^2+1 and I eventually get:

    int 3(u-1)+4/u^2 du/2 When I further break this down, I get:

    1/2 int 3u^-1 -u^-2 du am I on the right track? When I integrate this, I'm thinking that I have to be doing something wrong...

    Any suggestions?
     
  2. jcsd
  3. Jun 14, 2007 #2

    Kurdt

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    It looks fine to me as far as the final line there is a slight mistake with the signage.
     
  4. Jun 14, 2007 #3
    thanks for the 2nd set of eyes... Maybe it should read:

    1/2 int (3u^-1) - (u^-2) du

    I end up with 1/2 u^-1 +c or 1/2(x^2 + 1) + c since u=x^2+1
     
  5. Jun 14, 2007 #4

    Kurdt

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    You have:

    [tex]\int \frac{3x^3+4x}{(x^2+1)^2}dx[/tex]

    Put [tex] u=x^2+1 \Rightarrow du=2xdx[/tex]

    Which gives:

    [tex]\frac{3}{2} \int \frac{du}{u} - \frac{3}{2} \int \frac{du}{u^2} + 2 \int \frac{du}{u^2} [/tex]

    If you simplify that it will give you something slightly different to what you have. Its basically just a change of sign. Then all you need to do is evaluate the integrals.
     
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