Solving the indefinite integral $int frac{1}{x^2+x+1} dx$












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$begingroup$


I want to solve the following indefinite integral:
(1) $$int frac{1}{x^2+x+1} dx$$
completing the square:
(2) $$=int frac{1}{(x+frac{1}{2})^2+frac{3}{4}} dx$$
Substitution:
(3) $$u=frac{2x+1}{sqrt 3}$$ brings:
(4) $$dx=frac{sqrt 3}{2} du$$
(5) $$= frac{2}{sqrt 3}int frac{1}{u^2+1} du$$
Then getting standard integral
(6) $$= arctan(u)$$
and solving the integral and substitute back is not the problem. But I have a problem understanding the substitution in step (3). I don't know where the $sqrt 3$ beneath the fraction line comes from and how the formula ends in the standard integral in (5) on the right side. Could someone explain these steps in a bit more detail?






real-analysis integration indefinite-integrals fractions substitution







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  • $begingroup$
    Try $u=x+frac{1}{2}$ or $u=2x+1$ and see what happens. It may work it may not. Just a thought. Often proofs are worked out in a slightly ad-hoc and messy way, then the mathematician will tidy up his/her proof to present the most elegant form.
    $endgroup$
    – Antinous
    Jan 5 '18 at 9:17


















0












$begingroup$


I want to solve the following indefinite integral:
(1) $$int frac{1}{x^2+x+1} dx$$
completing the square:
(2) $$=int frac{1}{(x+frac{1}{2})^2+frac{3}{4}} dx$$
Substitution:
(3) $$u=frac{2x+1}{sqrt 3}$$ brings:
(4) $$dx=frac{sqrt 3}{2} du$$
(5) $$= frac{2}{sqrt 3}int frac{1}{u^2+1} du$$
Then getting standard integral
(6) $$= arctan(u)$$
and solving the integral and substitute back is not the problem. But I have a problem understanding the substitution in step (3). I don't know where the $sqrt 3$ beneath the fraction line comes from and how the formula ends in the standard integral in (5) on the right side. Could someone explain these steps in a bit more detail?






real-analysis integration indefinite-integrals fractions substitution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try $u=x+frac{1}{2}$ or $u=2x+1$ and see what happens. It may work it may not. Just a thought. Often proofs are worked out in a slightly ad-hoc and messy way, then the mathematician will tidy up his/her proof to present the most elegant form.
    $endgroup$
    – Antinous
    Jan 5 '18 at 9:17
















0












0








0





$begingroup$


I want to solve the following indefinite integral:
(1) $$int frac{1}{x^2+x+1} dx$$
completing the square:
(2) $$=int frac{1}{(x+frac{1}{2})^2+frac{3}{4}} dx$$
Substitution:
(3) $$u=frac{2x+1}{sqrt 3}$$ brings:
(4) $$dx=frac{sqrt 3}{2} du$$
(5) $$= frac{2}{sqrt 3}int frac{1}{u^2+1} du$$
Then getting standard integral
(6) $$= arctan(u)$$
and solving the integral and substitute back is not the problem. But I have a problem understanding the substitution in step (3). I don't know where the $sqrt 3$ beneath the fraction line comes from and how the formula ends in the standard integral in (5) on the right side. Could someone explain these steps in a bit more detail?






real-analysis integration indefinite-integrals fractions substitution







share|cite|improve this question











$endgroup$




I want to solve the following indefinite integral:
(1) $$int frac{1}{x^2+x+1} dx$$
completing the square:
(2) $$=int frac{1}{(x+frac{1}{2})^2+frac{3}{4}} dx$$
Substitution:
(3) $$u=frac{2x+1}{sqrt 3}$$ brings:
(4) $$dx=frac{sqrt 3}{2} du$$
(5) $$= frac{2}{sqrt 3}int frac{1}{u^2+1} du$$
Then getting standard integral
(6) $$= arctan(u)$$
and solving the integral and substitute back is not the problem. But I have a problem understanding the substitution in step (3). I don't know where the $sqrt 3$ beneath the fraction line comes from and how the formula ends in the standard integral in (5) on the right side. Could someone explain these steps in a bit more detail?






real-analysis integration indefinite-integrals fractions substitution




real-analysis integration indefinite-integrals fractions substitution






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edited Jan 5 '18 at 9:31









Michael Rozenberg

105k1892198




105k1892198










asked Jan 5 '18 at 9:13









MBDMBD

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  • $begingroup$
    Try $u=x+frac{1}{2}$ or $u=2x+1$ and see what happens. It may work it may not. Just a thought. Often proofs are worked out in a slightly ad-hoc and messy way, then the mathematician will tidy up his/her proof to present the most elegant form.
    $endgroup$
    – Antinous
    Jan 5 '18 at 9:17




















  • $begingroup$
    Try $u=x+frac{1}{2}$ or $u=2x+1$ and see what happens. It may work it may not. Just a thought. Often proofs are worked out in a slightly ad-hoc and messy way, then the mathematician will tidy up his/her proof to present the most elegant form.
    $endgroup$
    – Antinous
    Jan 5 '18 at 9:17


















$begingroup$
Try $u=x+frac{1}{2}$ or $u=2x+1$ and see what happens. It may work it may not. Just a thought. Often proofs are worked out in a slightly ad-hoc and messy way, then the mathematician will tidy up his/her proof to present the most elegant form.
$endgroup$
– Antinous
Jan 5 '18 at 9:17






$begingroup$
Try $u=x+frac{1}{2}$ or $u=2x+1$ and see what happens. It may work it may not. Just a thought. Often proofs are worked out in a slightly ad-hoc and messy way, then the mathematician will tidy up his/her proof to present the most elegant form.
$endgroup$
– Antinous
Jan 5 '18 at 9:17












2 Answers
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$begingroup$

$$left(x+frac{1}{2}right)^2+frac{3}{4}=left(frac{2x+1}{2}right)^2+left(frac{sqrt3}{2}right)^2=frac{3}{4}left(left(frac{2x+1}{sqrt3}right)^2+1right)=frac{3}{4}(u^2+1).$$
Since $frac{2x+1}{sqrt3}=u$, we have $$left(frac{2x+1}{sqrt3}right)'dx=du$$ or $$frac{2}{sqrt3}dx=du$$ or
$$dx=frac{sqrt3}{2}du.$$
Thus, $$intfrac{1}{x^2+x+1}dx=intfrac{1}{frac{3}{4}(u^2+1)}cdotfrac{sqrt3}{2}du=frac{2}{sqrt3}intfrac{1}{1+u^2}du.$$






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    $begingroup$

    We may take a more general approach here. Consider the integral
    $$I(x;a,b,c)=intfrac{mathrm dx}{ax^2+bx+c}=intfrac{mathrm dx}{a(x+frac{b}{2a})^2+g}$$
    Here $g=c-frac{b^2}{4a}$. If we assume that $4ac>b^2$, then we may make the substitution $x+frac{b}{2a}=sqrt{frac{g}{a}}tan u$ which gives
    $$I(x;a,b,c)=sqrt{frac{g}{a}}intfrac{sec^2u,mathrm du}{gtan^2u+g}$$
    $$I(x;a,b,c)=frac{u}{sqrt{ag}}$$
    $$I(x;a,b,c)=frac2{sqrt{4ac-b^2}}arctanfrac{2ax+b}{sqrt{4ac-b^2}}+C$$
    Plug in $a=b=c=1$ to get your integral






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      $begingroup$

      $$left(x+frac{1}{2}right)^2+frac{3}{4}=left(frac{2x+1}{2}right)^2+left(frac{sqrt3}{2}right)^2=frac{3}{4}left(left(frac{2x+1}{sqrt3}right)^2+1right)=frac{3}{4}(u^2+1).$$
      Since $frac{2x+1}{sqrt3}=u$, we have $$left(frac{2x+1}{sqrt3}right)'dx=du$$ or $$frac{2}{sqrt3}dx=du$$ or
      $$dx=frac{sqrt3}{2}du.$$
      Thus, $$intfrac{1}{x^2+x+1}dx=intfrac{1}{frac{3}{4}(u^2+1)}cdotfrac{sqrt3}{2}du=frac{2}{sqrt3}intfrac{1}{1+u^2}du.$$






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        $begingroup$

        $$left(x+frac{1}{2}right)^2+frac{3}{4}=left(frac{2x+1}{2}right)^2+left(frac{sqrt3}{2}right)^2=frac{3}{4}left(left(frac{2x+1}{sqrt3}right)^2+1right)=frac{3}{4}(u^2+1).$$
        Since $frac{2x+1}{sqrt3}=u$, we have $$left(frac{2x+1}{sqrt3}right)'dx=du$$ or $$frac{2}{sqrt3}dx=du$$ or
        $$dx=frac{sqrt3}{2}du.$$
        Thus, $$intfrac{1}{x^2+x+1}dx=intfrac{1}{frac{3}{4}(u^2+1)}cdotfrac{sqrt3}{2}du=frac{2}{sqrt3}intfrac{1}{1+u^2}du.$$






        share|cite|improve this answer











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          2












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          2





          $begingroup$

          $$left(x+frac{1}{2}right)^2+frac{3}{4}=left(frac{2x+1}{2}right)^2+left(frac{sqrt3}{2}right)^2=frac{3}{4}left(left(frac{2x+1}{sqrt3}right)^2+1right)=frac{3}{4}(u^2+1).$$
          Since $frac{2x+1}{sqrt3}=u$, we have $$left(frac{2x+1}{sqrt3}right)'dx=du$$ or $$frac{2}{sqrt3}dx=du$$ or
          $$dx=frac{sqrt3}{2}du.$$
          Thus, $$intfrac{1}{x^2+x+1}dx=intfrac{1}{frac{3}{4}(u^2+1)}cdotfrac{sqrt3}{2}du=frac{2}{sqrt3}intfrac{1}{1+u^2}du.$$






          share|cite|improve this answer











          $endgroup$



          $$left(x+frac{1}{2}right)^2+frac{3}{4}=left(frac{2x+1}{2}right)^2+left(frac{sqrt3}{2}right)^2=frac{3}{4}left(left(frac{2x+1}{sqrt3}right)^2+1right)=frac{3}{4}(u^2+1).$$
          Since $frac{2x+1}{sqrt3}=u$, we have $$left(frac{2x+1}{sqrt3}right)'dx=du$$ or $$frac{2}{sqrt3}dx=du$$ or
          $$dx=frac{sqrt3}{2}du.$$
          Thus, $$intfrac{1}{x^2+x+1}dx=intfrac{1}{frac{3}{4}(u^2+1)}cdotfrac{sqrt3}{2}du=frac{2}{sqrt3}intfrac{1}{1+u^2}du.$$







          share|cite|improve this answer














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          edited Jan 5 '18 at 9:23

























          answered Jan 5 '18 at 9:16









          Michael RozenbergMichael Rozenberg

          105k1892198




          105k1892198























              0












              $begingroup$

              We may take a more general approach here. Consider the integral
              $$I(x;a,b,c)=intfrac{mathrm dx}{ax^2+bx+c}=intfrac{mathrm dx}{a(x+frac{b}{2a})^2+g}$$
              Here $g=c-frac{b^2}{4a}$. If we assume that $4ac>b^2$, then we may make the substitution $x+frac{b}{2a}=sqrt{frac{g}{a}}tan u$ which gives
              $$I(x;a,b,c)=sqrt{frac{g}{a}}intfrac{sec^2u,mathrm du}{gtan^2u+g}$$
              $$I(x;a,b,c)=frac{u}{sqrt{ag}}$$
              $$I(x;a,b,c)=frac2{sqrt{4ac-b^2}}arctanfrac{2ax+b}{sqrt{4ac-b^2}}+C$$
              Plug in $a=b=c=1$ to get your integral






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                We may take a more general approach here. Consider the integral
                $$I(x;a,b,c)=intfrac{mathrm dx}{ax^2+bx+c}=intfrac{mathrm dx}{a(x+frac{b}{2a})^2+g}$$
                Here $g=c-frac{b^2}{4a}$. If we assume that $4ac>b^2$, then we may make the substitution $x+frac{b}{2a}=sqrt{frac{g}{a}}tan u$ which gives
                $$I(x;a,b,c)=sqrt{frac{g}{a}}intfrac{sec^2u,mathrm du}{gtan^2u+g}$$
                $$I(x;a,b,c)=frac{u}{sqrt{ag}}$$
                $$I(x;a,b,c)=frac2{sqrt{4ac-b^2}}arctanfrac{2ax+b}{sqrt{4ac-b^2}}+C$$
                Plug in $a=b=c=1$ to get your integral






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  We may take a more general approach here. Consider the integral
                  $$I(x;a,b,c)=intfrac{mathrm dx}{ax^2+bx+c}=intfrac{mathrm dx}{a(x+frac{b}{2a})^2+g}$$
                  Here $g=c-frac{b^2}{4a}$. If we assume that $4ac>b^2$, then we may make the substitution $x+frac{b}{2a}=sqrt{frac{g}{a}}tan u$ which gives
                  $$I(x;a,b,c)=sqrt{frac{g}{a}}intfrac{sec^2u,mathrm du}{gtan^2u+g}$$
                  $$I(x;a,b,c)=frac{u}{sqrt{ag}}$$
                  $$I(x;a,b,c)=frac2{sqrt{4ac-b^2}}arctanfrac{2ax+b}{sqrt{4ac-b^2}}+C$$
                  Plug in $a=b=c=1$ to get your integral






                  share|cite|improve this answer









                  $endgroup$



                  We may take a more general approach here. Consider the integral
                  $$I(x;a,b,c)=intfrac{mathrm dx}{ax^2+bx+c}=intfrac{mathrm dx}{a(x+frac{b}{2a})^2+g}$$
                  Here $g=c-frac{b^2}{4a}$. If we assume that $4ac>b^2$, then we may make the substitution $x+frac{b}{2a}=sqrt{frac{g}{a}}tan u$ which gives
                  $$I(x;a,b,c)=sqrt{frac{g}{a}}intfrac{sec^2u,mathrm du}{gtan^2u+g}$$
                  $$I(x;a,b,c)=frac{u}{sqrt{ag}}$$
                  $$I(x;a,b,c)=frac2{sqrt{4ac-b^2}}arctanfrac{2ax+b}{sqrt{4ac-b^2}}+C$$
                  Plug in $a=b=c=1$ to get your integral







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 22 at 4:39









                  clathratusclathratus

                  4,725337




                  4,725337






























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