Integration of rational function with substitution












0












$begingroup$


I am having trouble with integrating this function with substitution.



$
intfrac{x^2-4x}{x-2sqrt{x}}dx
$



First i do this:



$
u=sqrt{x}
$



$
du=frac{1}{2sqrt{x}}dx
$



$
dx=frac{du}{frac{1}{{2sqrt{x}}}}
$



then this:



$
intfrac{x^2-4x}{x-2u} frac{du}{frac{1}{{2sqrt{x}}}}
$



But now i don't know if what i have done is correct and know how to proceed.
Any help would be appreciated.






integration







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I am having trouble with integrating this function with substitution.



    $
    intfrac{x^2-4x}{x-2sqrt{x}}dx
    $



    First i do this:



    $
    u=sqrt{x}
    $



    $
    du=frac{1}{2sqrt{x}}dx
    $



    $
    dx=frac{du}{frac{1}{{2sqrt{x}}}}
    $



    then this:



    $
    intfrac{x^2-4x}{x-2u} frac{du}{frac{1}{{2sqrt{x}}}}
    $



    But now i don't know if what i have done is correct and know how to proceed.
    Any help would be appreciated.






    integration







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I am having trouble with integrating this function with substitution.



      $
      intfrac{x^2-4x}{x-2sqrt{x}}dx
      $



      First i do this:



      $
      u=sqrt{x}
      $



      $
      du=frac{1}{2sqrt{x}}dx
      $



      $
      dx=frac{du}{frac{1}{{2sqrt{x}}}}
      $



      then this:



      $
      intfrac{x^2-4x}{x-2u} frac{du}{frac{1}{{2sqrt{x}}}}
      $



      But now i don't know if what i have done is correct and know how to proceed.
      Any help would be appreciated.






      integration







      share|cite|improve this question









      $endgroup$




      I am having trouble with integrating this function with substitution.



      $
      intfrac{x^2-4x}{x-2sqrt{x}}dx
      $



      First i do this:



      $
      u=sqrt{x}
      $



      $
      du=frac{1}{2sqrt{x}}dx
      $



      $
      dx=frac{du}{frac{1}{{2sqrt{x}}}}
      $



      then this:



      $
      intfrac{x^2-4x}{x-2u} frac{du}{frac{1}{{2sqrt{x}}}}
      $



      But now i don't know if what i have done is correct and know how to proceed.
      Any help would be appreciated.






      integration




      integration






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Feb 1 at 20:13









      krnekikrneki

      82




      82






















          2 Answers
          2






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          2












          $begingroup$

          Note that $$(x-2sqrt{x})(x+2sqrt{x})=x^2-4x$$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            If you note that with your substitution follows $x=u^2$ and $x^2=u^4$ you get



            $int frac{x^2-4x}{x-2sqrt{x}}dx=2int u^2(u+2)du$



            for $u=sqrt{x}$ from which you should be able to continue.






            share|cite|improve this answer









            $endgroup$














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              2 Answers
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              2 Answers
              2






              active

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              active

              oldest

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              active

              oldest

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              2












              $begingroup$

              Note that $$(x-2sqrt{x})(x+2sqrt{x})=x^2-4x$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Note that $$(x-2sqrt{x})(x+2sqrt{x})=x^2-4x$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Note that $$(x-2sqrt{x})(x+2sqrt{x})=x^2-4x$$






                  share|cite|improve this answer









                  $endgroup$



                  Note that $$(x-2sqrt{x})(x+2sqrt{x})=x^2-4x$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 1 at 20:15









                  Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                  78.9k42867




                  78.9k42867























                      0












                      $begingroup$

                      If you note that with your substitution follows $x=u^2$ and $x^2=u^4$ you get



                      $int frac{x^2-4x}{x-2sqrt{x}}dx=2int u^2(u+2)du$



                      for $u=sqrt{x}$ from which you should be able to continue.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        If you note that with your substitution follows $x=u^2$ and $x^2=u^4$ you get



                        $int frac{x^2-4x}{x-2sqrt{x}}dx=2int u^2(u+2)du$



                        for $u=sqrt{x}$ from which you should be able to continue.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          If you note that with your substitution follows $x=u^2$ and $x^2=u^4$ you get



                          $int frac{x^2-4x}{x-2sqrt{x}}dx=2int u^2(u+2)du$



                          for $u=sqrt{x}$ from which you should be able to continue.






                          share|cite|improve this answer









                          $endgroup$



                          If you note that with your substitution follows $x=u^2$ and $x^2=u^4$ you get



                          $int frac{x^2-4x}{x-2sqrt{x}}dx=2int u^2(u+2)du$



                          for $u=sqrt{x}$ from which you should be able to continue.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Feb 1 at 20:54









                          babemcnuggetsbabemcnuggets

                          116110




                          116110






























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