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I'm trying find this integral

  1. Mar 27, 2005 #1
    Hi,

    I'm trying find this integral:

    [tex]
    \int \frac{dx}{\sqrt{x} \left(1+x^2\right)}
    [/tex]

    I tried the substitution [itex]t = \sqrt{x}[/itex] and I got

    [tex]
    2 \int \frac{dt}{1+t^4}
    [/tex]

    which I'm not able to solve. I know it should be some simple substitution, but I can't see any...

    Thank you for any help.
     
  2. jcsd
  3. Mar 27, 2005 #2
    Make your [itex] t= \sqrt{x}[/itex] substitution, then factor the denominator:

    [tex]1+t^4 = \left(t^2 + \frac{2}{\sqrt{2}}t + 1\right)\left(t^2 - \frac{2}{\sqrt{2}}t + 1\right)[/tex]

    then use partial fractions.
     
    Last edited: Mar 27, 2005
  4. Mar 27, 2005 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    U need partial fraction decomposition for the integral

    [tex] \int \frac{dt}{1+t^{4}} [/tex]

    HINT:

    [tex] \frac{1}{1+t^{4}}=\frac{1}{1+2t^{2}+t^{4}-2t^{2}}=\frac{1}{\left(1+t^{2}\right)^{2}-(x\sqrt{2})^{2}} [/tex]

    Now use that

    [tex] a^{2}-b^{2}\equiv (a+b)(a-b) [/tex]

    to get the partial fraction decomposition.

    In the end,u'll have to evaluate 4 simple integrals,2 involving artangent & 2 involving natural logarithm.

    Daniel.
     
  5. Mar 27, 2005 #4
    To add to what dexter and I have already said, remember that you can always factor a polynomial over the reals into quadractic and linear factors and apply partial fractions if necessary.
     
  6. Mar 27, 2005 #5
    That is wrong, though.

    [tex] x = \tan^2{t} \Longrightarrow x^2 = \tan^4{t} = \sin^4{t}\sec^4{t} \neq \sin^2{t}\sec^2{t}[/tex]
     
  7. Mar 27, 2005 #6

    dextercioby

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    Science Advisor
    Homework Helper

    What Data said it's true,but it's very rare that this decomposition/factoring is possible.The polyinomial that u had was really simple,but consider this one


    [tex]

    \int \frac{dx}{3x^7+2x^6-3x^5+2x^3+8}= [/tex]

    [tex]\sum_{R=\rho }R\ln \left( x+\frac{50537\,71425\,23884\,37683\,12969\,72723\,32288}{1\,47925\,11206\,35366\,91747\,99713}R^6+\frac{718\,89436\,36041\,51477\,11255\,98547\,18976}{1\,47925\,11206\,35366\,91747\,99713}R^5 [/tex]


    [tex]+\frac{12\,11818\,20513\,41769\,59090\,66072\,73984}{1\,47925\,11206\,35366\,91747\,99713}R^4+\frac{53395\,47640\,47920\,70339\,42785\,91360}{1\,47925\,11206\,35366\,91747\,99713}R^3+\frac{28\,14374\,51428\,61007\,71200\,20096}{1\,47925\,11206\,35366\,91747\,99713}R^2[/tex]

    [tex]\left +\frac{86\,45544\,28717\,76847\,41993\,27736}{1\,47925\,11206\,35366\,91747\,99713}R+\frac{48554\,15892\,19677\,88753\,18529}{4\,43775\,33619\,06100\,75243\,99139} \right) +C [/tex]

    [tex] \mbox{where}\ \rho \ \mbox {is a root of} \ 15418\,67541\,62688 x^7+6\,15789\,52704x^5 \\ +922226688x^4-196992x^3+257184x^2-1520x-243 [/tex]
     
    Last edited: Mar 27, 2005
  8. Mar 27, 2005 #7
    Yeah. It's always possible, but not easy in general :smile:
     
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