Solve $intfrac{2x-3}{(x^2+x+1)^2}dx$












2














$intfrac{2x-3}{(x^2+x+1)^2}dx$





$intfrac{2x-3}{(x^2+x+1)^2}dx=intfrac{2x+1}{(x^2+x+1)^2}dx-intfrac{4}{(x^2+x+1)^2}dx$


First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.










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    2














    $intfrac{2x-3}{(x^2+x+1)^2}dx$





    $intfrac{2x-3}{(x^2+x+1)^2}dx=intfrac{2x+1}{(x^2+x+1)^2}dx-intfrac{4}{(x^2+x+1)^2}dx$


    First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.










    share|cite|improve this question

























      2












      2








      2


      1





      $intfrac{2x-3}{(x^2+x+1)^2}dx$





      $intfrac{2x-3}{(x^2+x+1)^2}dx=intfrac{2x+1}{(x^2+x+1)^2}dx-intfrac{4}{(x^2+x+1)^2}dx$


      First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.










      share|cite|improve this question













      $intfrac{2x-3}{(x^2+x+1)^2}dx$





      $intfrac{2x-3}{(x^2+x+1)^2}dx=intfrac{2x+1}{(x^2+x+1)^2}dx-intfrac{4}{(x^2+x+1)^2}dx$


      First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.







      integration






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      asked Nov 23 at 13:34









      user984325

      17912




      17912






















          2 Answers
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          5














          Hint:



          As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$






          share|cite|improve this answer





























            3














            $$dfrac{dleft(dfrac{ax^2+bx+c}{x^2+x+1}right)}{dx}=dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$



            The numerator $(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)=x^2(a-b)+x(2a+2c)+b-c$



            If the numerator $2x-3,$



            $a-b=0iff a=b$



            $b-c=-3iff c=b+3$



            $2(a+c)=2iff1=a+c=b+b+3iff b=-1$






            share|cite|improve this answer





















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              2 Answers
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              active

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              2 Answers
              2






              active

              oldest

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              active

              oldest

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              active

              oldest

              votes









              5














              Hint:



              As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$






              share|cite|improve this answer


























                5














                Hint:



                As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$






                share|cite|improve this answer
























                  5












                  5








                  5






                  Hint:



                  As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$






                  share|cite|improve this answer












                  Hint:



                  As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 23 at 13:36









                  lab bhattacharjee

                  222k15155273




                  222k15155273























                      3














                      $$dfrac{dleft(dfrac{ax^2+bx+c}{x^2+x+1}right)}{dx}=dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$



                      The numerator $(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)=x^2(a-b)+x(2a+2c)+b-c$



                      If the numerator $2x-3,$



                      $a-b=0iff a=b$



                      $b-c=-3iff c=b+3$



                      $2(a+c)=2iff1=a+c=b+b+3iff b=-1$






                      share|cite|improve this answer


























                        3














                        $$dfrac{dleft(dfrac{ax^2+bx+c}{x^2+x+1}right)}{dx}=dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$



                        The numerator $(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)=x^2(a-b)+x(2a+2c)+b-c$



                        If the numerator $2x-3,$



                        $a-b=0iff a=b$



                        $b-c=-3iff c=b+3$



                        $2(a+c)=2iff1=a+c=b+b+3iff b=-1$






                        share|cite|improve this answer
























                          3












                          3








                          3






                          $$dfrac{dleft(dfrac{ax^2+bx+c}{x^2+x+1}right)}{dx}=dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$



                          The numerator $(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)=x^2(a-b)+x(2a+2c)+b-c$



                          If the numerator $2x-3,$



                          $a-b=0iff a=b$



                          $b-c=-3iff c=b+3$



                          $2(a+c)=2iff1=a+c=b+b+3iff b=-1$






                          share|cite|improve this answer












                          $$dfrac{dleft(dfrac{ax^2+bx+c}{x^2+x+1}right)}{dx}=dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$



                          The numerator $(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)=x^2(a-b)+x(2a+2c)+b-c$



                          If the numerator $2x-3,$



                          $a-b=0iff a=b$



                          $b-c=-3iff c=b+3$



                          $2(a+c)=2iff1=a+c=b+b+3iff b=-1$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 23 at 13:48









                          lab bhattacharjee

                          222k15155273




                          222k15155273






























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