How can I evaluate $int frac {x^2 + 1}{x^4 + x^2 +1} dx$ by partial fraction method?












0












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How can I evaluate $int frac {x^2 + 1}{x^4 + x^2 +1} dx$ by partial fraction method?



Could anyone tell me please?






calculus integration







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  • $begingroup$
    What do you know about partial fraction decomposition?
    $endgroup$
    – Arthur
    Dec 12 '18 at 10:03






  • 1




    $begingroup$
    Partial fraction decomposition provides an algorithm to rewrite your integrand. Have you tried to apply it?
    $endgroup$
    – Christoph
    Dec 12 '18 at 10:04
















0












$begingroup$


How can I evaluate $int frac {x^2 + 1}{x^4 + x^2 +1} dx$ by partial fraction method?



Could anyone tell me please?






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you know about partial fraction decomposition?
    $endgroup$
    – Arthur
    Dec 12 '18 at 10:03






  • 1




    $begingroup$
    Partial fraction decomposition provides an algorithm to rewrite your integrand. Have you tried to apply it?
    $endgroup$
    – Christoph
    Dec 12 '18 at 10:04














0












0








0





$begingroup$


How can I evaluate $int frac {x^2 + 1}{x^4 + x^2 +1} dx$ by partial fraction method?



Could anyone tell me please?






calculus integration







share|cite|improve this question











$endgroup$




How can I evaluate $int frac {x^2 + 1}{x^4 + x^2 +1} dx$ by partial fraction method?



Could anyone tell me please?






calculus integration




calculus integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 10:01









Arthur

121k7122208




121k7122208










asked Dec 12 '18 at 9:57









hopefullyhopefully

315215




315215












  • $begingroup$
    What do you know about partial fraction decomposition?
    $endgroup$
    – Arthur
    Dec 12 '18 at 10:03






  • 1




    $begingroup$
    Partial fraction decomposition provides an algorithm to rewrite your integrand. Have you tried to apply it?
    $endgroup$
    – Christoph
    Dec 12 '18 at 10:04


















  • $begingroup$
    What do you know about partial fraction decomposition?
    $endgroup$
    – Arthur
    Dec 12 '18 at 10:03






  • 1




    $begingroup$
    Partial fraction decomposition provides an algorithm to rewrite your integrand. Have you tried to apply it?
    $endgroup$
    – Christoph
    Dec 12 '18 at 10:04
















$begingroup$
What do you know about partial fraction decomposition?
$endgroup$
– Arthur
Dec 12 '18 at 10:03




$begingroup$
What do you know about partial fraction decomposition?
$endgroup$
– Arthur
Dec 12 '18 at 10:03




1




1




$begingroup$
Partial fraction decomposition provides an algorithm to rewrite your integrand. Have you tried to apply it?
$endgroup$
– Christoph
Dec 12 '18 at 10:04




$begingroup$
Partial fraction decomposition provides an algorithm to rewrite your integrand. Have you tried to apply it?
$endgroup$
– Christoph
Dec 12 '18 at 10:04










2 Answers
2






active

oldest

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3












$begingroup$

Hint: Notice that $1+x^2+x^4 = (1+x^2)^2-x^2=(1+x^2+x)(1+x^2-x)$, so your function can be rewritten as
begin{align}frac{A}{1+x+x^2}+frac{B}{1-x+x^2} & = frac{A-Ax+Ax^2+B+Bx+Bx^2}{1+x^2+x^4} \
& = frac{(A+B)+(B-A)x+(A+B)x^2}{1+x^2+x^4}
end{align}
and your conditions are $A+B=1$, $A-B = 0$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Not what you're asking, but here's a different method



    $$ int frac{x^2+1}{x^4+x^2+1}dx = int frac{1 + frac{1}{x^2}}{x^2 + frac{1}{x^2} + 1}dx = int frac{1 + frac{1}{x^2}}{left(x - frac{1}{x}right)^2 + 3}dx $$



    Let $u = x-frac{1}{x}$, then the integral becomes



    $$ int frac{1}{u^2+3}du = frac{1}{sqrt{3}}arctanfrac{u}{sqrt{3}} + C $$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Hint: Notice that $1+x^2+x^4 = (1+x^2)^2-x^2=(1+x^2+x)(1+x^2-x)$, so your function can be rewritten as
      begin{align}frac{A}{1+x+x^2}+frac{B}{1-x+x^2} & = frac{A-Ax+Ax^2+B+Bx+Bx^2}{1+x^2+x^4} \
      & = frac{(A+B)+(B-A)x+(A+B)x^2}{1+x^2+x^4}
      end{align}
      and your conditions are $A+B=1$, $A-B = 0$.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Hint: Notice that $1+x^2+x^4 = (1+x^2)^2-x^2=(1+x^2+x)(1+x^2-x)$, so your function can be rewritten as
        begin{align}frac{A}{1+x+x^2}+frac{B}{1-x+x^2} & = frac{A-Ax+Ax^2+B+Bx+Bx^2}{1+x^2+x^4} \
        & = frac{(A+B)+(B-A)x+(A+B)x^2}{1+x^2+x^4}
        end{align}
        and your conditions are $A+B=1$, $A-B = 0$.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Hint: Notice that $1+x^2+x^4 = (1+x^2)^2-x^2=(1+x^2+x)(1+x^2-x)$, so your function can be rewritten as
          begin{align}frac{A}{1+x+x^2}+frac{B}{1-x+x^2} & = frac{A-Ax+Ax^2+B+Bx+Bx^2}{1+x^2+x^4} \
          & = frac{(A+B)+(B-A)x+(A+B)x^2}{1+x^2+x^4}
          end{align}
          and your conditions are $A+B=1$, $A-B = 0$.






          share|cite|improve this answer









          $endgroup$



          Hint: Notice that $1+x^2+x^4 = (1+x^2)^2-x^2=(1+x^2+x)(1+x^2-x)$, so your function can be rewritten as
          begin{align}frac{A}{1+x+x^2}+frac{B}{1-x+x^2} & = frac{A-Ax+Ax^2+B+Bx+Bx^2}{1+x^2+x^4} \
          & = frac{(A+B)+(B-A)x+(A+B)x^2}{1+x^2+x^4}
          end{align}
          and your conditions are $A+B=1$, $A-B = 0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 10:04









          GibbsGibbs

          5,4383927




          5,4383927























              1












              $begingroup$

              Not what you're asking, but here's a different method



              $$ int frac{x^2+1}{x^4+x^2+1}dx = int frac{1 + frac{1}{x^2}}{x^2 + frac{1}{x^2} + 1}dx = int frac{1 + frac{1}{x^2}}{left(x - frac{1}{x}right)^2 + 3}dx $$



              Let $u = x-frac{1}{x}$, then the integral becomes



              $$ int frac{1}{u^2+3}du = frac{1}{sqrt{3}}arctanfrac{u}{sqrt{3}} + C $$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Not what you're asking, but here's a different method



                $$ int frac{x^2+1}{x^4+x^2+1}dx = int frac{1 + frac{1}{x^2}}{x^2 + frac{1}{x^2} + 1}dx = int frac{1 + frac{1}{x^2}}{left(x - frac{1}{x}right)^2 + 3}dx $$



                Let $u = x-frac{1}{x}$, then the integral becomes



                $$ int frac{1}{u^2+3}du = frac{1}{sqrt{3}}arctanfrac{u}{sqrt{3}} + C $$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Not what you're asking, but here's a different method



                  $$ int frac{x^2+1}{x^4+x^2+1}dx = int frac{1 + frac{1}{x^2}}{x^2 + frac{1}{x^2} + 1}dx = int frac{1 + frac{1}{x^2}}{left(x - frac{1}{x}right)^2 + 3}dx $$



                  Let $u = x-frac{1}{x}$, then the integral becomes



                  $$ int frac{1}{u^2+3}du = frac{1}{sqrt{3}}arctanfrac{u}{sqrt{3}} + C $$






                  share|cite|improve this answer









                  $endgroup$



                  Not what you're asking, but here's a different method



                  $$ int frac{x^2+1}{x^4+x^2+1}dx = int frac{1 + frac{1}{x^2}}{x^2 + frac{1}{x^2} + 1}dx = int frac{1 + frac{1}{x^2}}{left(x - frac{1}{x}right)^2 + 3}dx $$



                  Let $u = x-frac{1}{x}$, then the integral becomes



                  $$ int frac{1}{u^2+3}du = frac{1}{sqrt{3}}arctanfrac{u}{sqrt{3}} + C $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 10:54









                  DylanDylan

                  14.2k31127




                  14.2k31127






























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