Integrating $int (t^2+7)^{frac{1}{3}}$dt












1












$begingroup$



I have to integrate $int (t^2+7)^{frac{1}{3}}$dt




I tried some trigonometric substitution and then rationalisation but didn't get anything



Please provide a hint on how to proceed with this question



Please help!!!






calculus integration indefinite-integrals hypergeometric-function







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I don't think you can expect an elementary anti-derivative. Are you sure you're supposed to find one (by hand)?
    $endgroup$
    – StackTD
    Jan 24 '18 at 15:29










  • $begingroup$
    Are you sure about the power $frac 13$ ? This makes the problem very difficult (no closed form solution). I bet for $frac 12$.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 15:30










  • $begingroup$
    No it is 1/3 for sure
    $endgroup$
    – Atul Mishra
    Jan 24 '18 at 15:30










  • $begingroup$
    Do you want to bet ? I am sure that, once more, there is a typo in a textbook.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 15:31










  • $begingroup$
    Any answer is appreciated@StackTD
    $endgroup$
    – Atul Mishra
    Jan 24 '18 at 15:32
















1












$begingroup$



I have to integrate $int (t^2+7)^{frac{1}{3}}$dt




I tried some trigonometric substitution and then rationalisation but didn't get anything



Please provide a hint on how to proceed with this question



Please help!!!






calculus integration indefinite-integrals hypergeometric-function







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I don't think you can expect an elementary anti-derivative. Are you sure you're supposed to find one (by hand)?
    $endgroup$
    – StackTD
    Jan 24 '18 at 15:29










  • $begingroup$
    Are you sure about the power $frac 13$ ? This makes the problem very difficult (no closed form solution). I bet for $frac 12$.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 15:30










  • $begingroup$
    No it is 1/3 for sure
    $endgroup$
    – Atul Mishra
    Jan 24 '18 at 15:30










  • $begingroup$
    Do you want to bet ? I am sure that, once more, there is a typo in a textbook.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 15:31










  • $begingroup$
    Any answer is appreciated@StackTD
    $endgroup$
    – Atul Mishra
    Jan 24 '18 at 15:32














1












1








1


2



$begingroup$



I have to integrate $int (t^2+7)^{frac{1}{3}}$dt




I tried some trigonometric substitution and then rationalisation but didn't get anything



Please provide a hint on how to proceed with this question



Please help!!!






calculus integration indefinite-integrals hypergeometric-function







share|cite|improve this question











$endgroup$





I have to integrate $int (t^2+7)^{frac{1}{3}}$dt




I tried some trigonometric substitution and then rationalisation but didn't get anything



Please provide a hint on how to proceed with this question



Please help!!!






calculus integration indefinite-integrals hypergeometric-function




calculus integration indefinite-integrals hypergeometric-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 '18 at 15:48









Guy Fsone

17.3k43074




17.3k43074










asked Jan 24 '18 at 15:26









Atul MishraAtul Mishra

1,8991031




1,8991031








  • 3




    $begingroup$
    I don't think you can expect an elementary anti-derivative. Are you sure you're supposed to find one (by hand)?
    $endgroup$
    – StackTD
    Jan 24 '18 at 15:29










  • $begingroup$
    Are you sure about the power $frac 13$ ? This makes the problem very difficult (no closed form solution). I bet for $frac 12$.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 15:30










  • $begingroup$
    No it is 1/3 for sure
    $endgroup$
    – Atul Mishra
    Jan 24 '18 at 15:30










  • $begingroup$
    Do you want to bet ? I am sure that, once more, there is a typo in a textbook.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 15:31










  • $begingroup$
    Any answer is appreciated@StackTD
    $endgroup$
    – Atul Mishra
    Jan 24 '18 at 15:32














  • 3




    $begingroup$
    I don't think you can expect an elementary anti-derivative. Are you sure you're supposed to find one (by hand)?
    $endgroup$
    – StackTD
    Jan 24 '18 at 15:29










  • $begingroup$
    Are you sure about the power $frac 13$ ? This makes the problem very difficult (no closed form solution). I bet for $frac 12$.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 15:30










  • $begingroup$
    No it is 1/3 for sure
    $endgroup$
    – Atul Mishra
    Jan 24 '18 at 15:30










  • $begingroup$
    Do you want to bet ? I am sure that, once more, there is a typo in a textbook.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 15:31










  • $begingroup$
    Any answer is appreciated@StackTD
    $endgroup$
    – Atul Mishra
    Jan 24 '18 at 15:32








3




3




$begingroup$
I don't think you can expect an elementary anti-derivative. Are you sure you're supposed to find one (by hand)?
$endgroup$
– StackTD
Jan 24 '18 at 15:29




$begingroup$
I don't think you can expect an elementary anti-derivative. Are you sure you're supposed to find one (by hand)?
$endgroup$
– StackTD
Jan 24 '18 at 15:29












$begingroup$
Are you sure about the power $frac 13$ ? This makes the problem very difficult (no closed form solution). I bet for $frac 12$.
$endgroup$
– Claude Leibovici
Jan 24 '18 at 15:30




$begingroup$
Are you sure about the power $frac 13$ ? This makes the problem very difficult (no closed form solution). I bet for $frac 12$.
$endgroup$
– Claude Leibovici
Jan 24 '18 at 15:30












$begingroup$
No it is 1/3 for sure
$endgroup$
– Atul Mishra
Jan 24 '18 at 15:30




$begingroup$
No it is 1/3 for sure
$endgroup$
– Atul Mishra
Jan 24 '18 at 15:30












$begingroup$
Do you want to bet ? I am sure that, once more, there is a typo in a textbook.
$endgroup$
– Claude Leibovici
Jan 24 '18 at 15:31




$begingroup$
Do you want to bet ? I am sure that, once more, there is a typo in a textbook.
$endgroup$
– Claude Leibovici
Jan 24 '18 at 15:31












$begingroup$
Any answer is appreciated@StackTD
$endgroup$
– Atul Mishra
Jan 24 '18 at 15:32




$begingroup$
Any answer is appreciated@StackTD
$endgroup$
– Atul Mishra
Jan 24 '18 at 15:32










2 Answers
2






active

oldest

votes


















3












$begingroup$

Setting $$t=sqrt{7}cosh ximplies dt =sqrt{7}sinh xdx$$



$$int (t^2+7)^{frac{1}{3}}dt = 7^{2/3}int (cosh^2x+1)^{frac{1}{3}} sinh x, dx = 7^{2/3}int sinh^{5/3} x, dx$$



this integral cannot be written as elementary function. It belongs to the class of hypergeometric functions






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I tried that , but power is 1/3 , this is not helping for 1/3 power
    $endgroup$
    – Atul Mishra
    Jan 24 '18 at 15:35










  • $begingroup$
    I bet this is the hint for exponent $frac12$.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 15:35






  • 1




    $begingroup$
    This is very fine for $frac 12$ but $frac 13$ leads to hypergeometric function, I guess.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 15:36








  • 1




    $begingroup$
    @ClaudeLeibovici ya I just check it does not an obvious integral
    $endgroup$
    – Guy Fsone
    Jan 24 '18 at 15:37






  • 1




    $begingroup$
    @Isham it will make it worse
    $endgroup$
    – Guy Fsone
    Jan 24 '18 at 15:42



















0












$begingroup$

Positioning your integral:



begin{equation}
int left(t^2 + 7 right)^{frac{1}{3}}:dt = int_0^t left(w^2 + 7 right)^{frac{1}{3}}:dw = int_0^t frac{1}{ left(w^2 + 7 right)^{frac{1}{3}}}:dw
end{equation}



We can employ the solution that I've spoken to here:
begin{equation}
int_0^x frac{t^k}{left(t^n + aright)^m}:dt = frac{1}{n}a^{frac{k + 1}{n} - m} left[Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right) - Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}, frac{1}{1 + ax^n} right)right]
end{equation}



Here $a = 7$, $m = -frac{1}{3}$, $k = 0$, and $n = 2$. Thus,



begin{align}
int left(t^2 + 7 right)^{frac{1}{3}}:dt &= frac{1}{2}cdot 7^{frac{0 + 1}{2} --frac{1}{3}}left[Bleft(-frac{1}{3} - frac{0 + 1}{2}, frac{0 + 1}{2}right) - Bleft(-frac{1}{3}- frac{0 + 1}{2}, frac{0 + 1}{2}, frac{1}{1 + 7t^2} right)right] \
&= frac{7^{frac{5}{6}}}{2}left[Bleft(-frac{5}{6}, frac{1}{2}right) - Bleft(-frac{5}{6}, frac{1}{2}, frac{1}{1 + 7t^2} right)right]
end{align}






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












    $begingroup$

    Setting $$t=sqrt{7}cosh ximplies dt =sqrt{7}sinh xdx$$



    $$int (t^2+7)^{frac{1}{3}}dt = 7^{2/3}int (cosh^2x+1)^{frac{1}{3}} sinh x, dx = 7^{2/3}int sinh^{5/3} x, dx$$



    this integral cannot be written as elementary function. It belongs to the class of hypergeometric functions






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I tried that , but power is 1/3 , this is not helping for 1/3 power
      $endgroup$
      – Atul Mishra
      Jan 24 '18 at 15:35










    • $begingroup$
      I bet this is the hint for exponent $frac12$.
      $endgroup$
      – Weijun Zhou
      Jan 24 '18 at 15:35






    • 1




      $begingroup$
      This is very fine for $frac 12$ but $frac 13$ leads to hypergeometric function, I guess.
      $endgroup$
      – Claude Leibovici
      Jan 24 '18 at 15:36








    • 1




      $begingroup$
      @ClaudeLeibovici ya I just check it does not an obvious integral
      $endgroup$
      – Guy Fsone
      Jan 24 '18 at 15:37






    • 1




      $begingroup$
      @Isham it will make it worse
      $endgroup$
      – Guy Fsone
      Jan 24 '18 at 15:42
















    3












    $begingroup$

    Setting $$t=sqrt{7}cosh ximplies dt =sqrt{7}sinh xdx$$



    $$int (t^2+7)^{frac{1}{3}}dt = 7^{2/3}int (cosh^2x+1)^{frac{1}{3}} sinh x, dx = 7^{2/3}int sinh^{5/3} x, dx$$



    this integral cannot be written as elementary function. It belongs to the class of hypergeometric functions






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I tried that , but power is 1/3 , this is not helping for 1/3 power
      $endgroup$
      – Atul Mishra
      Jan 24 '18 at 15:35










    • $begingroup$
      I bet this is the hint for exponent $frac12$.
      $endgroup$
      – Weijun Zhou
      Jan 24 '18 at 15:35






    • 1




      $begingroup$
      This is very fine for $frac 12$ but $frac 13$ leads to hypergeometric function, I guess.
      $endgroup$
      – Claude Leibovici
      Jan 24 '18 at 15:36








    • 1




      $begingroup$
      @ClaudeLeibovici ya I just check it does not an obvious integral
      $endgroup$
      – Guy Fsone
      Jan 24 '18 at 15:37






    • 1




      $begingroup$
      @Isham it will make it worse
      $endgroup$
      – Guy Fsone
      Jan 24 '18 at 15:42














    3












    3








    3





    $begingroup$

    Setting $$t=sqrt{7}cosh ximplies dt =sqrt{7}sinh xdx$$



    $$int (t^2+7)^{frac{1}{3}}dt = 7^{2/3}int (cosh^2x+1)^{frac{1}{3}} sinh x, dx = 7^{2/3}int sinh^{5/3} x, dx$$



    this integral cannot be written as elementary function. It belongs to the class of hypergeometric functions






    share|cite|improve this answer











    $endgroup$



    Setting $$t=sqrt{7}cosh ximplies dt =sqrt{7}sinh xdx$$



    $$int (t^2+7)^{frac{1}{3}}dt = 7^{2/3}int (cosh^2x+1)^{frac{1}{3}} sinh x, dx = 7^{2/3}int sinh^{5/3} x, dx$$



    this integral cannot be written as elementary function. It belongs to the class of hypergeometric functions







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 24 '18 at 15:45

























    answered Jan 24 '18 at 15:33









    Guy FsoneGuy Fsone

    17.3k43074




    17.3k43074












    • $begingroup$
      I tried that , but power is 1/3 , this is not helping for 1/3 power
      $endgroup$
      – Atul Mishra
      Jan 24 '18 at 15:35










    • $begingroup$
      I bet this is the hint for exponent $frac12$.
      $endgroup$
      – Weijun Zhou
      Jan 24 '18 at 15:35






    • 1




      $begingroup$
      This is very fine for $frac 12$ but $frac 13$ leads to hypergeometric function, I guess.
      $endgroup$
      – Claude Leibovici
      Jan 24 '18 at 15:36








    • 1




      $begingroup$
      @ClaudeLeibovici ya I just check it does not an obvious integral
      $endgroup$
      – Guy Fsone
      Jan 24 '18 at 15:37






    • 1




      $begingroup$
      @Isham it will make it worse
      $endgroup$
      – Guy Fsone
      Jan 24 '18 at 15:42


















    • $begingroup$
      I tried that , but power is 1/3 , this is not helping for 1/3 power
      $endgroup$
      – Atul Mishra
      Jan 24 '18 at 15:35










    • $begingroup$
      I bet this is the hint for exponent $frac12$.
      $endgroup$
      – Weijun Zhou
      Jan 24 '18 at 15:35






    • 1




      $begingroup$
      This is very fine for $frac 12$ but $frac 13$ leads to hypergeometric function, I guess.
      $endgroup$
      – Claude Leibovici
      Jan 24 '18 at 15:36








    • 1




      $begingroup$
      @ClaudeLeibovici ya I just check it does not an obvious integral
      $endgroup$
      – Guy Fsone
      Jan 24 '18 at 15:37






    • 1




      $begingroup$
      @Isham it will make it worse
      $endgroup$
      – Guy Fsone
      Jan 24 '18 at 15:42
















    $begingroup$
    I tried that , but power is 1/3 , this is not helping for 1/3 power
    $endgroup$
    – Atul Mishra
    Jan 24 '18 at 15:35




    $begingroup$
    I tried that , but power is 1/3 , this is not helping for 1/3 power
    $endgroup$
    – Atul Mishra
    Jan 24 '18 at 15:35












    $begingroup$
    I bet this is the hint for exponent $frac12$.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 15:35




    $begingroup$
    I bet this is the hint for exponent $frac12$.
    $endgroup$
    – Weijun Zhou
    Jan 24 '18 at 15:35




    1




    1




    $begingroup$
    This is very fine for $frac 12$ but $frac 13$ leads to hypergeometric function, I guess.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 15:36






    $begingroup$
    This is very fine for $frac 12$ but $frac 13$ leads to hypergeometric function, I guess.
    $endgroup$
    – Claude Leibovici
    Jan 24 '18 at 15:36






    1




    1




    $begingroup$
    @ClaudeLeibovici ya I just check it does not an obvious integral
    $endgroup$
    – Guy Fsone
    Jan 24 '18 at 15:37




    $begingroup$
    @ClaudeLeibovici ya I just check it does not an obvious integral
    $endgroup$
    – Guy Fsone
    Jan 24 '18 at 15:37




    1




    1




    $begingroup$
    @Isham it will make it worse
    $endgroup$
    – Guy Fsone
    Jan 24 '18 at 15:42




    $begingroup$
    @Isham it will make it worse
    $endgroup$
    – Guy Fsone
    Jan 24 '18 at 15:42











    0












    $begingroup$

    Positioning your integral:



    begin{equation}
    int left(t^2 + 7 right)^{frac{1}{3}}:dt = int_0^t left(w^2 + 7 right)^{frac{1}{3}}:dw = int_0^t frac{1}{ left(w^2 + 7 right)^{frac{1}{3}}}:dw
    end{equation}



    We can employ the solution that I've spoken to here:
    begin{equation}
    int_0^x frac{t^k}{left(t^n + aright)^m}:dt = frac{1}{n}a^{frac{k + 1}{n} - m} left[Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right) - Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}, frac{1}{1 + ax^n} right)right]
    end{equation}



    Here $a = 7$, $m = -frac{1}{3}$, $k = 0$, and $n = 2$. Thus,



    begin{align}
    int left(t^2 + 7 right)^{frac{1}{3}}:dt &= frac{1}{2}cdot 7^{frac{0 + 1}{2} --frac{1}{3}}left[Bleft(-frac{1}{3} - frac{0 + 1}{2}, frac{0 + 1}{2}right) - Bleft(-frac{1}{3}- frac{0 + 1}{2}, frac{0 + 1}{2}, frac{1}{1 + 7t^2} right)right] \
    &= frac{7^{frac{5}{6}}}{2}left[Bleft(-frac{5}{6}, frac{1}{2}right) - Bleft(-frac{5}{6}, frac{1}{2}, frac{1}{1 + 7t^2} right)right]
    end{align}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Positioning your integral:



      begin{equation}
      int left(t^2 + 7 right)^{frac{1}{3}}:dt = int_0^t left(w^2 + 7 right)^{frac{1}{3}}:dw = int_0^t frac{1}{ left(w^2 + 7 right)^{frac{1}{3}}}:dw
      end{equation}



      We can employ the solution that I've spoken to here:
      begin{equation}
      int_0^x frac{t^k}{left(t^n + aright)^m}:dt = frac{1}{n}a^{frac{k + 1}{n} - m} left[Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right) - Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}, frac{1}{1 + ax^n} right)right]
      end{equation}



      Here $a = 7$, $m = -frac{1}{3}$, $k = 0$, and $n = 2$. Thus,



      begin{align}
      int left(t^2 + 7 right)^{frac{1}{3}}:dt &= frac{1}{2}cdot 7^{frac{0 + 1}{2} --frac{1}{3}}left[Bleft(-frac{1}{3} - frac{0 + 1}{2}, frac{0 + 1}{2}right) - Bleft(-frac{1}{3}- frac{0 + 1}{2}, frac{0 + 1}{2}, frac{1}{1 + 7t^2} right)right] \
      &= frac{7^{frac{5}{6}}}{2}left[Bleft(-frac{5}{6}, frac{1}{2}right) - Bleft(-frac{5}{6}, frac{1}{2}, frac{1}{1 + 7t^2} right)right]
      end{align}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Positioning your integral:



        begin{equation}
        int left(t^2 + 7 right)^{frac{1}{3}}:dt = int_0^t left(w^2 + 7 right)^{frac{1}{3}}:dw = int_0^t frac{1}{ left(w^2 + 7 right)^{frac{1}{3}}}:dw
        end{equation}



        We can employ the solution that I've spoken to here:
        begin{equation}
        int_0^x frac{t^k}{left(t^n + aright)^m}:dt = frac{1}{n}a^{frac{k + 1}{n} - m} left[Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right) - Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}, frac{1}{1 + ax^n} right)right]
        end{equation}



        Here $a = 7$, $m = -frac{1}{3}$, $k = 0$, and $n = 2$. Thus,



        begin{align}
        int left(t^2 + 7 right)^{frac{1}{3}}:dt &= frac{1}{2}cdot 7^{frac{0 + 1}{2} --frac{1}{3}}left[Bleft(-frac{1}{3} - frac{0 + 1}{2}, frac{0 + 1}{2}right) - Bleft(-frac{1}{3}- frac{0 + 1}{2}, frac{0 + 1}{2}, frac{1}{1 + 7t^2} right)right] \
        &= frac{7^{frac{5}{6}}}{2}left[Bleft(-frac{5}{6}, frac{1}{2}right) - Bleft(-frac{5}{6}, frac{1}{2}, frac{1}{1 + 7t^2} right)right]
        end{align}






        share|cite|improve this answer









        $endgroup$



        Positioning your integral:



        begin{equation}
        int left(t^2 + 7 right)^{frac{1}{3}}:dt = int_0^t left(w^2 + 7 right)^{frac{1}{3}}:dw = int_0^t frac{1}{ left(w^2 + 7 right)^{frac{1}{3}}}:dw
        end{equation}



        We can employ the solution that I've spoken to here:
        begin{equation}
        int_0^x frac{t^k}{left(t^n + aright)^m}:dt = frac{1}{n}a^{frac{k + 1}{n} - m} left[Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right) - Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}, frac{1}{1 + ax^n} right)right]
        end{equation}



        Here $a = 7$, $m = -frac{1}{3}$, $k = 0$, and $n = 2$. Thus,



        begin{align}
        int left(t^2 + 7 right)^{frac{1}{3}}:dt &= frac{1}{2}cdot 7^{frac{0 + 1}{2} --frac{1}{3}}left[Bleft(-frac{1}{3} - frac{0 + 1}{2}, frac{0 + 1}{2}right) - Bleft(-frac{1}{3}- frac{0 + 1}{2}, frac{0 + 1}{2}, frac{1}{1 + 7t^2} right)right] \
        &= frac{7^{frac{5}{6}}}{2}left[Bleft(-frac{5}{6}, frac{1}{2}right) - Bleft(-frac{5}{6}, frac{1}{2}, frac{1}{1 + 7t^2} right)right]
        end{align}







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        answered Dec 31 '18 at 2:12







        user150203





































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