How to solve $int :frac3x^2+1x^3+2dx$?

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I am trying to solve the following integral:



$$int :frac3x^2+1x^3+2dx$$



I am not sure how to do it. I tried using partial fractions but you can not factor $x^3+2$, and changing the variable with $u$ does not seem to work either. Anyone has any ideas?




calculus integration indefinite-integrals




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  • The easy part is $frac3x^2x^3+2$ indeed $(log x^3+2)'=frac3x^2x^3+2$.
    – gimusi
    Jul 23 at 21:02










  • But I guess that does not help at all because the remaining integral would be $ int frac1x^3+2mathrmdx$ and I am not sure how to solve this.
    – mrtaurho
    Jul 23 at 21:03







  • 3




    You can factor any polynomial over $mathbbR$ into linear or 2nd degree polynomials. Here,you can factor $x^3+2=(x+sqrt[3] 2)(x^2-sqrt[3]2x+sqrt[3]4)$, if its any help.
    – Sar
    Jul 23 at 21:04










  • Separate the two and in the second integral take sub $1+2/x^3=t$
    – Pi_die_die
    Jul 23 at 21:13










  • This is it ! You can combine gimusi's trick and Dr. Sonnhard Graubner's answer. The linear and quadratic factors are mutually prime so you get $$frac1x^3+2=fracAx+sqrt[3]2+fracBx+Cx^2-sqrt[3]2x+2^2/3$$ the second can be written $$fraclambda Q'Q+fracmuQ$$ where $$Q(x)=x^2-sqrt[3]2x+2^2/3$$
    – Duchamp Gérard H. E.
    Jul 23 at 21:19














up vote
0
down vote

favorite
1












I am trying to solve the following integral:



$$int :frac3x^2+1x^3+2dx$$



I am not sure how to do it. I tried using partial fractions but you can not factor $x^3+2$, and changing the variable with $u$ does not seem to work either. Anyone has any ideas?




calculus integration indefinite-integrals




share|cite|improve this question





















  • The easy part is $frac3x^2x^3+2$ indeed $(log x^3+2)'=frac3x^2x^3+2$.
    – gimusi
    Jul 23 at 21:02










  • But I guess that does not help at all because the remaining integral would be $ int frac1x^3+2mathrmdx$ and I am not sure how to solve this.
    – mrtaurho
    Jul 23 at 21:03







  • 3




    You can factor any polynomial over $mathbbR$ into linear or 2nd degree polynomials. Here,you can factor $x^3+2=(x+sqrt[3] 2)(x^2-sqrt[3]2x+sqrt[3]4)$, if its any help.
    – Sar
    Jul 23 at 21:04










  • Separate the two and in the second integral take sub $1+2/x^3=t$
    – Pi_die_die
    Jul 23 at 21:13










  • This is it ! You can combine gimusi's trick and Dr. Sonnhard Graubner's answer. The linear and quadratic factors are mutually prime so you get $$frac1x^3+2=fracAx+sqrt[3]2+fracBx+Cx^2-sqrt[3]2x+2^2/3$$ the second can be written $$fraclambda Q'Q+fracmuQ$$ where $$Q(x)=x^2-sqrt[3]2x+2^2/3$$
    – Duchamp Gérard H. E.
    Jul 23 at 21:19












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





I am trying to solve the following integral:



$$int :frac3x^2+1x^3+2dx$$



I am not sure how to do it. I tried using partial fractions but you can not factor $x^3+2$, and changing the variable with $u$ does not seem to work either. Anyone has any ideas?




calculus integration indefinite-integrals




share|cite|improve this question













I am trying to solve the following integral:



$$int :frac3x^2+1x^3+2dx$$



I am not sure how to do it. I tried using partial fractions but you can not factor $x^3+2$, and changing the variable with $u$ does not seem to work either. Anyone has any ideas?





calculus integration indefinite-integrals





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 23 at 21:00









Abcd

2,3361624




2,3361624









asked Jul 23 at 20:56









Francis

543




543











  • The easy part is $frac3x^2x^3+2$ indeed $(log x^3+2)'=frac3x^2x^3+2$.
    – gimusi
    Jul 23 at 21:02










  • But I guess that does not help at all because the remaining integral would be $ int frac1x^3+2mathrmdx$ and I am not sure how to solve this.
    – mrtaurho
    Jul 23 at 21:03







  • 3




    You can factor any polynomial over $mathbbR$ into linear or 2nd degree polynomials. Here,you can factor $x^3+2=(x+sqrt[3] 2)(x^2-sqrt[3]2x+sqrt[3]4)$, if its any help.
    – Sar
    Jul 23 at 21:04










  • Separate the two and in the second integral take sub $1+2/x^3=t$
    – Pi_die_die
    Jul 23 at 21:13










  • This is it ! You can combine gimusi's trick and Dr. Sonnhard Graubner's answer. The linear and quadratic factors are mutually prime so you get $$frac1x^3+2=fracAx+sqrt[3]2+fracBx+Cx^2-sqrt[3]2x+2^2/3$$ the second can be written $$fraclambda Q'Q+fracmuQ$$ where $$Q(x)=x^2-sqrt[3]2x+2^2/3$$
    – Duchamp Gérard H. E.
    Jul 23 at 21:19
















  • The easy part is $frac3x^2x^3+2$ indeed $(log x^3+2)'=frac3x^2x^3+2$.
    – gimusi
    Jul 23 at 21:02










  • But I guess that does not help at all because the remaining integral would be $ int frac1x^3+2mathrmdx$ and I am not sure how to solve this.
    – mrtaurho
    Jul 23 at 21:03







  • 3




    You can factor any polynomial over $mathbbR$ into linear or 2nd degree polynomials. Here,you can factor $x^3+2=(x+sqrt[3] 2)(x^2-sqrt[3]2x+sqrt[3]4)$, if its any help.
    – Sar
    Jul 23 at 21:04










  • Separate the two and in the second integral take sub $1+2/x^3=t$
    – Pi_die_die
    Jul 23 at 21:13










  • This is it ! You can combine gimusi's trick and Dr. Sonnhard Graubner's answer. The linear and quadratic factors are mutually prime so you get $$frac1x^3+2=fracAx+sqrt[3]2+fracBx+Cx^2-sqrt[3]2x+2^2/3$$ the second can be written $$fraclambda Q'Q+fracmuQ$$ where $$Q(x)=x^2-sqrt[3]2x+2^2/3$$
    – Duchamp Gérard H. E.
    Jul 23 at 21:19















The easy part is $frac3x^2x^3+2$ indeed $(log x^3+2)'=frac3x^2x^3+2$.
– gimusi
Jul 23 at 21:02




The easy part is $frac3x^2x^3+2$ indeed $(log x^3+2)'=frac3x^2x^3+2$.
– gimusi
Jul 23 at 21:02












But I guess that does not help at all because the remaining integral would be $ int frac1x^3+2mathrmdx$ and I am not sure how to solve this.
– mrtaurho
Jul 23 at 21:03





But I guess that does not help at all because the remaining integral would be $ int frac1x^3+2mathrmdx$ and I am not sure how to solve this.
– mrtaurho
Jul 23 at 21:03





3




3




You can factor any polynomial over $mathbbR$ into linear or 2nd degree polynomials. Here,you can factor $x^3+2=(x+sqrt[3] 2)(x^2-sqrt[3]2x+sqrt[3]4)$, if its any help.
– Sar
Jul 23 at 21:04




You can factor any polynomial over $mathbbR$ into linear or 2nd degree polynomials. Here,you can factor $x^3+2=(x+sqrt[3] 2)(x^2-sqrt[3]2x+sqrt[3]4)$, if its any help.
– Sar
Jul 23 at 21:04












Separate the two and in the second integral take sub $1+2/x^3=t$
– Pi_die_die
Jul 23 at 21:13




Separate the two and in the second integral take sub $1+2/x^3=t$
– Pi_die_die
Jul 23 at 21:13












This is it ! You can combine gimusi's trick and Dr. Sonnhard Graubner's answer. The linear and quadratic factors are mutually prime so you get $$frac1x^3+2=fracAx+sqrt[3]2+fracBx+Cx^2-sqrt[3]2x+2^2/3$$ the second can be written $$fraclambda Q'Q+fracmuQ$$ where $$Q(x)=x^2-sqrt[3]2x+2^2/3$$
– Duchamp Gérard H. E.
Jul 23 at 21:19




This is it ! You can combine gimusi's trick and Dr. Sonnhard Graubner's answer. The linear and quadratic factors are mutually prime so you get $$frac1x^3+2=fracAx+sqrt[3]2+fracBx+Cx^2-sqrt[3]2x+2^2/3$$ the second can be written $$fraclambda Q'Q+fracmuQ$$ where $$Q(x)=x^2-sqrt[3]2x+2^2/3$$
– Duchamp Gérard H. E.
Jul 23 at 21:19










4 Answers
4






active

oldest

votes

















up vote
0
down vote



accepted










Hint: write your integral in the form



$$int frac3x^2+1(x+sqrt[3]2)(x^2-sqrt[3]2x+2^2/3)dx$$






share|cite|improve this answer





















  • I guess you could apply partial fractions to this "new" but it seems to me that this gets really messy. Is there no other way?
    – mrtaurho
    Jul 23 at 21:11










  • I think here in this case is no other way, let me say i know no other way.
    – Dr. Sonnhard Graubner
    Jul 23 at 21:14










  • Yes. Technically you can factor anything and apply partial fractions in the real numbers, but I was trying to avoid this. I suppose there truly is no other way. Thanks for your help!
    – Francis
    Jul 23 at 21:15







  • 1




    You can clean things up a little by doing a substitution $u = frac1sqrt[3]2x$ first; doesn't help a lot, but it makes the numbers better.
    – Reese
    Jul 23 at 21:19

















up vote
1
down vote













Wolfy gives
a number of forms,
the first of which is



$ log(x^3 + 2) - dfraclog(2^1/3 x^2 - 2^2/3 x + 2)6 2^2/3
+ dfraclog(2^2/3 x + 2)3 2^2/3
+ dfractan^-1(frac2^2/3 x - 1sqrt3)2^2/3 sqrt3
+C
$






share|cite|improve this answer




























    up vote
    0
    down vote













    Substitute $u=x^3$ and you can reduce it down to:



    $$frac13intfracu^-frac23u+2du+intfrac1u+2du$$



    I believe it can be taken from here






    share|cite|improve this answer




























      up vote
      0
      down vote













      $$I=int frac3x^2+1x^3+2dx=int frac3x^2x^3+2dx+int fracdxx^3+2$$
      $$I=log(x^3+2)+int fracdxx^3+2$$ For the remaining integral, let
      $$x=sqrt[3]2, timplies dx=sqrt[3]2,dt$$
      $$J=int fracdxx^3+2=fracsqrt[3]22 int fracdtt^3+1$$
      Now
      $$t^3+1=(t-1)(t-a)(t-b)$$ Partial fraction decomposition
      $$frac 1 t^3+1=frac1(a-1) (a-b) (t-a)+frac1(b-1) (b-a) (t-b)+frac1(a-1) (b-1) (t-1)$$ Integrate to get a weighted sum of logarithms with complex values (the weights are complex too) and recombine them to get the answer already given.






      share|cite|improve this answer





















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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        0
        down vote



        accepted










        Hint: write your integral in the form



        $$int frac3x^2+1(x+sqrt[3]2)(x^2-sqrt[3]2x+2^2/3)dx$$






        share|cite|improve this answer





















        • I guess you could apply partial fractions to this "new" but it seems to me that this gets really messy. Is there no other way?
          – mrtaurho
          Jul 23 at 21:11










        • I think here in this case is no other way, let me say i know no other way.
          – Dr. Sonnhard Graubner
          Jul 23 at 21:14










        • Yes. Technically you can factor anything and apply partial fractions in the real numbers, but I was trying to avoid this. I suppose there truly is no other way. Thanks for your help!
          – Francis
          Jul 23 at 21:15







        • 1




          You can clean things up a little by doing a substitution $u = frac1sqrt[3]2x$ first; doesn't help a lot, but it makes the numbers better.
          – Reese
          Jul 23 at 21:19














        up vote
        0
        down vote



        accepted










        Hint: write your integral in the form



        $$int frac3x^2+1(x+sqrt[3]2)(x^2-sqrt[3]2x+2^2/3)dx$$






        share|cite|improve this answer





















        • I guess you could apply partial fractions to this "new" but it seems to me that this gets really messy. Is there no other way?
          – mrtaurho
          Jul 23 at 21:11










        • I think here in this case is no other way, let me say i know no other way.
          – Dr. Sonnhard Graubner
          Jul 23 at 21:14










        • Yes. Technically you can factor anything and apply partial fractions in the real numbers, but I was trying to avoid this. I suppose there truly is no other way. Thanks for your help!
          – Francis
          Jul 23 at 21:15







        • 1




          You can clean things up a little by doing a substitution $u = frac1sqrt[3]2x$ first; doesn't help a lot, but it makes the numbers better.
          – Reese
          Jul 23 at 21:19












        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        Hint: write your integral in the form



        $$int frac3x^2+1(x+sqrt[3]2)(x^2-sqrt[3]2x+2^2/3)dx$$






        share|cite|improve this answer













        Hint: write your integral in the form



        $$int frac3x^2+1(x+sqrt[3]2)(x^2-sqrt[3]2x+2^2/3)dx$$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 23 at 21:05









        Dr. Sonnhard Graubner

        66.7k32659




        66.7k32659











        • I guess you could apply partial fractions to this "new" but it seems to me that this gets really messy. Is there no other way?
          – mrtaurho
          Jul 23 at 21:11










        • I think here in this case is no other way, let me say i know no other way.
          – Dr. Sonnhard Graubner
          Jul 23 at 21:14










        • Yes. Technically you can factor anything and apply partial fractions in the real numbers, but I was trying to avoid this. I suppose there truly is no other way. Thanks for your help!
          – Francis
          Jul 23 at 21:15







        • 1




          You can clean things up a little by doing a substitution $u = frac1sqrt[3]2x$ first; doesn't help a lot, but it makes the numbers better.
          – Reese
          Jul 23 at 21:19
















        • I guess you could apply partial fractions to this "new" but it seems to me that this gets really messy. Is there no other way?
          – mrtaurho
          Jul 23 at 21:11










        • I think here in this case is no other way, let me say i know no other way.
          – Dr. Sonnhard Graubner
          Jul 23 at 21:14










        • Yes. Technically you can factor anything and apply partial fractions in the real numbers, but I was trying to avoid this. I suppose there truly is no other way. Thanks for your help!
          – Francis
          Jul 23 at 21:15







        • 1




          You can clean things up a little by doing a substitution $u = frac1sqrt[3]2x$ first; doesn't help a lot, but it makes the numbers better.
          – Reese
          Jul 23 at 21:19















        I guess you could apply partial fractions to this "new" but it seems to me that this gets really messy. Is there no other way?
        – mrtaurho
        Jul 23 at 21:11




        I guess you could apply partial fractions to this "new" but it seems to me that this gets really messy. Is there no other way?
        – mrtaurho
        Jul 23 at 21:11












        I think here in this case is no other way, let me say i know no other way.
        – Dr. Sonnhard Graubner
        Jul 23 at 21:14




        I think here in this case is no other way, let me say i know no other way.
        – Dr. Sonnhard Graubner
        Jul 23 at 21:14












        Yes. Technically you can factor anything and apply partial fractions in the real numbers, but I was trying to avoid this. I suppose there truly is no other way. Thanks for your help!
        – Francis
        Jul 23 at 21:15





        Yes. Technically you can factor anything and apply partial fractions in the real numbers, but I was trying to avoid this. I suppose there truly is no other way. Thanks for your help!
        – Francis
        Jul 23 at 21:15





        1




        1




        You can clean things up a little by doing a substitution $u = frac1sqrt[3]2x$ first; doesn't help a lot, but it makes the numbers better.
        – Reese
        Jul 23 at 21:19




        You can clean things up a little by doing a substitution $u = frac1sqrt[3]2x$ first; doesn't help a lot, but it makes the numbers better.
        – Reese
        Jul 23 at 21:19










        up vote
        1
        down vote













        Wolfy gives
        a number of forms,
        the first of which is



        $ log(x^3 + 2) - dfraclog(2^1/3 x^2 - 2^2/3 x + 2)6 2^2/3
        + dfraclog(2^2/3 x + 2)3 2^2/3
        + dfractan^-1(frac2^2/3 x - 1sqrt3)2^2/3 sqrt3
        +C
        $






        share|cite|improve this answer

























          up vote
          1
          down vote













          Wolfy gives
          a number of forms,
          the first of which is



          $ log(x^3 + 2) - dfraclog(2^1/3 x^2 - 2^2/3 x + 2)6 2^2/3
          + dfraclog(2^2/3 x + 2)3 2^2/3
          + dfractan^-1(frac2^2/3 x - 1sqrt3)2^2/3 sqrt3
          +C
          $






          share|cite|improve this answer























            up vote
            1
            down vote










            up vote
            1
            down vote









            Wolfy gives
            a number of forms,
            the first of which is



            $ log(x^3 + 2) - dfraclog(2^1/3 x^2 - 2^2/3 x + 2)6 2^2/3
            + dfraclog(2^2/3 x + 2)3 2^2/3
            + dfractan^-1(frac2^2/3 x - 1sqrt3)2^2/3 sqrt3
            +C
            $






            share|cite|improve this answer













            Wolfy gives
            a number of forms,
            the first of which is



            $ log(x^3 + 2) - dfraclog(2^1/3 x^2 - 2^2/3 x + 2)6 2^2/3
            + dfraclog(2^2/3 x + 2)3 2^2/3
            + dfractan^-1(frac2^2/3 x - 1sqrt3)2^2/3 sqrt3
            +C
            $







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered Jul 23 at 21:46









            marty cohen

            69.1k446122




            69.1k446122




















                up vote
                0
                down vote













                Substitute $u=x^3$ and you can reduce it down to:



                $$frac13intfracu^-frac23u+2du+intfrac1u+2du$$



                I believe it can be taken from here






                share|cite|improve this answer

























                  up vote
                  0
                  down vote













                  Substitute $u=x^3$ and you can reduce it down to:



                  $$frac13intfracu^-frac23u+2du+intfrac1u+2du$$



                  I believe it can be taken from here






                  share|cite|improve this answer























                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    Substitute $u=x^3$ and you can reduce it down to:



                    $$frac13intfracu^-frac23u+2du+intfrac1u+2du$$



                    I believe it can be taken from here






                    share|cite|improve this answer













                    Substitute $u=x^3$ and you can reduce it down to:



                    $$frac13intfracu^-frac23u+2du+intfrac1u+2du$$



                    I believe it can be taken from here







                    share|cite|improve this answer













                    share|cite|improve this answer



                    share|cite|improve this answer











                    answered Jul 23 at 21:19









                    Rhys Hughes

                    3,8681227




                    3,8681227




















                        up vote
                        0
                        down vote













                        $$I=int frac3x^2+1x^3+2dx=int frac3x^2x^3+2dx+int fracdxx^3+2$$
                        $$I=log(x^3+2)+int fracdxx^3+2$$ For the remaining integral, let
                        $$x=sqrt[3]2, timplies dx=sqrt[3]2,dt$$
                        $$J=int fracdxx^3+2=fracsqrt[3]22 int fracdtt^3+1$$
                        Now
                        $$t^3+1=(t-1)(t-a)(t-b)$$ Partial fraction decomposition
                        $$frac 1 t^3+1=frac1(a-1) (a-b) (t-a)+frac1(b-1) (b-a) (t-b)+frac1(a-1) (b-1) (t-1)$$ Integrate to get a weighted sum of logarithms with complex values (the weights are complex too) and recombine them to get the answer already given.






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          $$I=int frac3x^2+1x^3+2dx=int frac3x^2x^3+2dx+int fracdxx^3+2$$
                          $$I=log(x^3+2)+int fracdxx^3+2$$ For the remaining integral, let
                          $$x=sqrt[3]2, timplies dx=sqrt[3]2,dt$$
                          $$J=int fracdxx^3+2=fracsqrt[3]22 int fracdtt^3+1$$
                          Now
                          $$t^3+1=(t-1)(t-a)(t-b)$$ Partial fraction decomposition
                          $$frac 1 t^3+1=frac1(a-1) (a-b) (t-a)+frac1(b-1) (b-a) (t-b)+frac1(a-1) (b-1) (t-1)$$ Integrate to get a weighted sum of logarithms with complex values (the weights are complex too) and recombine them to get the answer already given.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            $$I=int frac3x^2+1x^3+2dx=int frac3x^2x^3+2dx+int fracdxx^3+2$$
                            $$I=log(x^3+2)+int fracdxx^3+2$$ For the remaining integral, let
                            $$x=sqrt[3]2, timplies dx=sqrt[3]2,dt$$
                            $$J=int fracdxx^3+2=fracsqrt[3]22 int fracdtt^3+1$$
                            Now
                            $$t^3+1=(t-1)(t-a)(t-b)$$ Partial fraction decomposition
                            $$frac 1 t^3+1=frac1(a-1) (a-b) (t-a)+frac1(b-1) (b-a) (t-b)+frac1(a-1) (b-1) (t-1)$$ Integrate to get a weighted sum of logarithms with complex values (the weights are complex too) and recombine them to get the answer already given.






                            share|cite|improve this answer













                            $$I=int frac3x^2+1x^3+2dx=int frac3x^2x^3+2dx+int fracdxx^3+2$$
                            $$I=log(x^3+2)+int fracdxx^3+2$$ For the remaining integral, let
                            $$x=sqrt[3]2, timplies dx=sqrt[3]2,dt$$
                            $$J=int fracdxx^3+2=fracsqrt[3]22 int fracdtt^3+1$$
                            Now
                            $$t^3+1=(t-1)(t-a)(t-b)$$ Partial fraction decomposition
                            $$frac 1 t^3+1=frac1(a-1) (a-b) (t-a)+frac1(b-1) (b-a) (t-b)+frac1(a-1) (b-1) (t-1)$$ Integrate to get a weighted sum of logarithms with complex values (the weights are complex too) and recombine them to get the answer already given.







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                            answered Jul 24 at 5:30









                            Claude Leibovici

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