Prove that $ intlimits_0^inftyfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},dx=dfrac{3pi}{8} $ [closed]












1












$begingroup$


Is this conjecture true?




Conjecture:
$$
int_0^inftyfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},dx=dfrac{3pi}{8}
$$




I found it myself based on numerical evidence. Need help in analytical proof. Thanks.










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closed as off-topic by Nosrati, Did, Zacky, RRL, amWhy Jan 5 at 21:19


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Did, Zacky, RRL, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    To the person who voted to close, care to explain???
    $endgroup$
    – Tyrell
    Jan 5 at 12:34








  • 2




    $begingroup$
    What have you tried sofar?
    $endgroup$
    – Nosrati
    Jan 5 at 12:35






  • 2




    $begingroup$
    @Nosrati what have I tried? First I found it by numerical calculation. I'm not very strong at analytics. I don't even know where to start tbh.
    $endgroup$
    – Tyrell
    Jan 5 at 12:37






  • 1




    $begingroup$
    Maybe you lack context on this question and that's why you got some close votes + downvotes? It is okay if you don't have any approaches, but how did this integral arrive to you? Surely you didn't dream about it. Also maybe you can include what are you studying :)
    $endgroup$
    – Zacky
    Jan 5 at 12:43






  • 1




    $begingroup$
    this might work: the integrant is even, so write it as an integral over the real numbers, find the zeros of the denominator, then evaluate the corresponding contour integrals.
    $endgroup$
    – john
    Jan 5 at 12:44
















1












$begingroup$


Is this conjecture true?




Conjecture:
$$
int_0^inftyfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},dx=dfrac{3pi}{8}
$$




I found it myself based on numerical evidence. Need help in analytical proof. Thanks.










share|cite|improve this question











$endgroup$



closed as off-topic by Nosrati, Did, Zacky, RRL, amWhy Jan 5 at 21:19


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Did, Zacky, RRL, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    To the person who voted to close, care to explain???
    $endgroup$
    – Tyrell
    Jan 5 at 12:34








  • 2




    $begingroup$
    What have you tried sofar?
    $endgroup$
    – Nosrati
    Jan 5 at 12:35






  • 2




    $begingroup$
    @Nosrati what have I tried? First I found it by numerical calculation. I'm not very strong at analytics. I don't even know where to start tbh.
    $endgroup$
    – Tyrell
    Jan 5 at 12:37






  • 1




    $begingroup$
    Maybe you lack context on this question and that's why you got some close votes + downvotes? It is okay if you don't have any approaches, but how did this integral arrive to you? Surely you didn't dream about it. Also maybe you can include what are you studying :)
    $endgroup$
    – Zacky
    Jan 5 at 12:43






  • 1




    $begingroup$
    this might work: the integrant is even, so write it as an integral over the real numbers, find the zeros of the denominator, then evaluate the corresponding contour integrals.
    $endgroup$
    – john
    Jan 5 at 12:44














1












1








1


1



$begingroup$


Is this conjecture true?




Conjecture:
$$
int_0^inftyfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},dx=dfrac{3pi}{8}
$$




I found it myself based on numerical evidence. Need help in analytical proof. Thanks.










share|cite|improve this question











$endgroup$




Is this conjecture true?




Conjecture:
$$
int_0^inftyfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},dx=dfrac{3pi}{8}
$$




I found it myself based on numerical evidence. Need help in analytical proof. Thanks.







integration definite-integrals closed-form






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share|cite|improve this question













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share|cite|improve this question








edited Jan 5 at 12:47









mrtaurho

5,70551540




5,70551540










asked Jan 5 at 12:26









TyrellTyrell

582321




582321




closed as off-topic by Nosrati, Did, Zacky, RRL, amWhy Jan 5 at 21:19


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Did, Zacky, RRL, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Nosrati, Did, Zacky, RRL, amWhy Jan 5 at 21:19


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Did, Zacky, RRL, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    To the person who voted to close, care to explain???
    $endgroup$
    – Tyrell
    Jan 5 at 12:34








  • 2




    $begingroup$
    What have you tried sofar?
    $endgroup$
    – Nosrati
    Jan 5 at 12:35






  • 2




    $begingroup$
    @Nosrati what have I tried? First I found it by numerical calculation. I'm not very strong at analytics. I don't even know where to start tbh.
    $endgroup$
    – Tyrell
    Jan 5 at 12:37






  • 1




    $begingroup$
    Maybe you lack context on this question and that's why you got some close votes + downvotes? It is okay if you don't have any approaches, but how did this integral arrive to you? Surely you didn't dream about it. Also maybe you can include what are you studying :)
    $endgroup$
    – Zacky
    Jan 5 at 12:43






  • 1




    $begingroup$
    this might work: the integrant is even, so write it as an integral over the real numbers, find the zeros of the denominator, then evaluate the corresponding contour integrals.
    $endgroup$
    – john
    Jan 5 at 12:44














  • 2




    $begingroup$
    To the person who voted to close, care to explain???
    $endgroup$
    – Tyrell
    Jan 5 at 12:34








  • 2




    $begingroup$
    What have you tried sofar?
    $endgroup$
    – Nosrati
    Jan 5 at 12:35






  • 2




    $begingroup$
    @Nosrati what have I tried? First I found it by numerical calculation. I'm not very strong at analytics. I don't even know where to start tbh.
    $endgroup$
    – Tyrell
    Jan 5 at 12:37






  • 1




    $begingroup$
    Maybe you lack context on this question and that's why you got some close votes + downvotes? It is okay if you don't have any approaches, but how did this integral arrive to you? Surely you didn't dream about it. Also maybe you can include what are you studying :)
    $endgroup$
    – Zacky
    Jan 5 at 12:43






  • 1




    $begingroup$
    this might work: the integrant is even, so write it as an integral over the real numbers, find the zeros of the denominator, then evaluate the corresponding contour integrals.
    $endgroup$
    – john
    Jan 5 at 12:44








2




2




$begingroup$
To the person who voted to close, care to explain???
$endgroup$
– Tyrell
Jan 5 at 12:34






$begingroup$
To the person who voted to close, care to explain???
$endgroup$
– Tyrell
Jan 5 at 12:34






2




2




$begingroup$
What have you tried sofar?
$endgroup$
– Nosrati
Jan 5 at 12:35




$begingroup$
What have you tried sofar?
$endgroup$
– Nosrati
Jan 5 at 12:35




2




2




$begingroup$
@Nosrati what have I tried? First I found it by numerical calculation. I'm not very strong at analytics. I don't even know where to start tbh.
$endgroup$
– Tyrell
Jan 5 at 12:37




$begingroup$
@Nosrati what have I tried? First I found it by numerical calculation. I'm not very strong at analytics. I don't even know where to start tbh.
$endgroup$
– Tyrell
Jan 5 at 12:37




1




1




$begingroup$
Maybe you lack context on this question and that's why you got some close votes + downvotes? It is okay if you don't have any approaches, but how did this integral arrive to you? Surely you didn't dream about it. Also maybe you can include what are you studying :)
$endgroup$
– Zacky
Jan 5 at 12:43




$begingroup$
Maybe you lack context on this question and that's why you got some close votes + downvotes? It is okay if you don't have any approaches, but how did this integral arrive to you? Surely you didn't dream about it. Also maybe you can include what are you studying :)
$endgroup$
– Zacky
Jan 5 at 12:43




1




1




$begingroup$
this might work: the integrant is even, so write it as an integral over the real numbers, find the zeros of the denominator, then evaluate the corresponding contour integrals.
$endgroup$
– john
Jan 5 at 12:44




$begingroup$
this might work: the integrant is even, so write it as an integral over the real numbers, find the zeros of the denominator, then evaluate the corresponding contour integrals.
$endgroup$
– john
Jan 5 at 12:44










3 Answers
3






active

oldest

votes


















4












$begingroup$

It is easier than I exspected in the first place. However, first of all lets denote your integral as




$$mathfrak I~=~int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x$$




We may rewrite the integral a little bit. According to Trebor's answer we have to exspect an inverse tangent with an argument of the form $z=e^{2x}frac{1-x}{1+x}$ thus write $mathfrak I$ as



$$begin{align*}
mathfrak I=int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x&=int_0^infty frac1{1+left(e^{2x}frac{1-x}{1+x}right)^2}frac{x^2e^{2x}mathrm d x}{(1+x)^2}
end{align*}$$



Now we can try the straightforward substitution $z=e^{2x}frac{1-x}{1+x}$ with a corresponding differential given by $dz=-2frac{x^2e^{2x}mathrm d x}{(1+x)^2}$ and therefore the whole integral boils down to



$$begin{align*}
mathfrak I &=int_0^infty frac1{1+left(e^{2x}frac{1-x}{1+x}right)^2}frac{x^2e^{2x}mathrm d x}{(1+x)^2}\
&=int_1^{-infty} frac1{1+z^2}frac{-mathrm dz}{2}\
&=frac12int_{-infty}^1 frac{mathrm d z}{1+z^2}\
&=frac12left[arctan{z}right]_{-infty}^1
end{align*}$$




$$therefore~mathfrak I ~=~int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x~=~frac{3pi}8$$




Where we used the symmetry aswell as well-known values of the tangent function.






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    6












    $begingroup$

    Well, we can just rewrite the integral as: $$int_0^inftyfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},dx=int_0^infty frac{x^2}{e^{2x}left(frac{1-x}{1+x}right)^2+e^{-2x}}frac{1}{(1+x)^2}dx$$
    $$=int_0^infty frac{1}{e^{4x}left(frac{1-x}{1+x}right)^2+1}frac{1}{e^{-2x}}frac{x^2}{(1+x)^2}dx =int_0^infty frac{color{blue}{e^{2x}}}{left(color{red}{e^{2x}cdotfrac{1-x}{1+x}}right)^2+1}color{blue}{frac{x^2}{(1+x)^2}dx}$$ Now set $,displaystyle{color{red}{e^{2x} frac{1-x}{1+x}}=tRightarrow color{blue}{e^{2x}frac{x^2}{(1+x)^2}dx}=-frac12 dt},$ and the result follows.






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      The integral is $$int_0^xfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},mathrm dx=frac{1}{8} left(pi +4 tan ^{-1}left(frac{(x-1) e^{2x}}{x+1}right)right)$$ as can be verified by direct differentiation.



      You can first tackle the $e^{2x}$ part by substitution. The rest should be a trivial matter.






      share|cite|improve this answer









      $endgroup$









      • 4




        $begingroup$
        How did you do this?
        $endgroup$
        – Tyrell
        Jan 5 at 12:44


















      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      It is easier than I exspected in the first place. However, first of all lets denote your integral as




      $$mathfrak I~=~int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x$$




      We may rewrite the integral a little bit. According to Trebor's answer we have to exspect an inverse tangent with an argument of the form $z=e^{2x}frac{1-x}{1+x}$ thus write $mathfrak I$ as



      $$begin{align*}
      mathfrak I=int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x&=int_0^infty frac1{1+left(e^{2x}frac{1-x}{1+x}right)^2}frac{x^2e^{2x}mathrm d x}{(1+x)^2}
      end{align*}$$



      Now we can try the straightforward substitution $z=e^{2x}frac{1-x}{1+x}$ with a corresponding differential given by $dz=-2frac{x^2e^{2x}mathrm d x}{(1+x)^2}$ and therefore the whole integral boils down to



      $$begin{align*}
      mathfrak I &=int_0^infty frac1{1+left(e^{2x}frac{1-x}{1+x}right)^2}frac{x^2e^{2x}mathrm d x}{(1+x)^2}\
      &=int_1^{-infty} frac1{1+z^2}frac{-mathrm dz}{2}\
      &=frac12int_{-infty}^1 frac{mathrm d z}{1+z^2}\
      &=frac12left[arctan{z}right]_{-infty}^1
      end{align*}$$




      $$therefore~mathfrak I ~=~int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x~=~frac{3pi}8$$




      Where we used the symmetry aswell as well-known values of the tangent function.






      share|cite|improve this answer











      $endgroup$


















        4












        $begingroup$

        It is easier than I exspected in the first place. However, first of all lets denote your integral as




        $$mathfrak I~=~int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x$$




        We may rewrite the integral a little bit. According to Trebor's answer we have to exspect an inverse tangent with an argument of the form $z=e^{2x}frac{1-x}{1+x}$ thus write $mathfrak I$ as



        $$begin{align*}
        mathfrak I=int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x&=int_0^infty frac1{1+left(e^{2x}frac{1-x}{1+x}right)^2}frac{x^2e^{2x}mathrm d x}{(1+x)^2}
        end{align*}$$



        Now we can try the straightforward substitution $z=e^{2x}frac{1-x}{1+x}$ with a corresponding differential given by $dz=-2frac{x^2e^{2x}mathrm d x}{(1+x)^2}$ and therefore the whole integral boils down to



        $$begin{align*}
        mathfrak I &=int_0^infty frac1{1+left(e^{2x}frac{1-x}{1+x}right)^2}frac{x^2e^{2x}mathrm d x}{(1+x)^2}\
        &=int_1^{-infty} frac1{1+z^2}frac{-mathrm dz}{2}\
        &=frac12int_{-infty}^1 frac{mathrm d z}{1+z^2}\
        &=frac12left[arctan{z}right]_{-infty}^1
        end{align*}$$




        $$therefore~mathfrak I ~=~int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x~=~frac{3pi}8$$




        Where we used the symmetry aswell as well-known values of the tangent function.






        share|cite|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$

          It is easier than I exspected in the first place. However, first of all lets denote your integral as




          $$mathfrak I~=~int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x$$




          We may rewrite the integral a little bit. According to Trebor's answer we have to exspect an inverse tangent with an argument of the form $z=e^{2x}frac{1-x}{1+x}$ thus write $mathfrak I$ as



          $$begin{align*}
          mathfrak I=int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x&=int_0^infty frac1{1+left(e^{2x}frac{1-x}{1+x}right)^2}frac{x^2e^{2x}mathrm d x}{(1+x)^2}
          end{align*}$$



          Now we can try the straightforward substitution $z=e^{2x}frac{1-x}{1+x}$ with a corresponding differential given by $dz=-2frac{x^2e^{2x}mathrm d x}{(1+x)^2}$ and therefore the whole integral boils down to



          $$begin{align*}
          mathfrak I &=int_0^infty frac1{1+left(e^{2x}frac{1-x}{1+x}right)^2}frac{x^2e^{2x}mathrm d x}{(1+x)^2}\
          &=int_1^{-infty} frac1{1+z^2}frac{-mathrm dz}{2}\
          &=frac12int_{-infty}^1 frac{mathrm d z}{1+z^2}\
          &=frac12left[arctan{z}right]_{-infty}^1
          end{align*}$$




          $$therefore~mathfrak I ~=~int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x~=~frac{3pi}8$$




          Where we used the symmetry aswell as well-known values of the tangent function.






          share|cite|improve this answer











          $endgroup$



          It is easier than I exspected in the first place. However, first of all lets denote your integral as




          $$mathfrak I~=~int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x$$




          We may rewrite the integral a little bit. According to Trebor's answer we have to exspect an inverse tangent with an argument of the form $z=e^{2x}frac{1-x}{1+x}$ thus write $mathfrak I$ as



          $$begin{align*}
          mathfrak I=int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x&=int_0^infty frac1{1+left(e^{2x}frac{1-x}{1+x}right)^2}frac{x^2e^{2x}mathrm d x}{(1+x)^2}
          end{align*}$$



          Now we can try the straightforward substitution $z=e^{2x}frac{1-x}{1+x}$ with a corresponding differential given by $dz=-2frac{x^2e^{2x}mathrm d x}{(1+x)^2}$ and therefore the whole integral boils down to



          $$begin{align*}
          mathfrak I &=int_0^infty frac1{1+left(e^{2x}frac{1-x}{1+x}right)^2}frac{x^2e^{2x}mathrm d x}{(1+x)^2}\
          &=int_1^{-infty} frac1{1+z^2}frac{-mathrm dz}{2}\
          &=frac12int_{-infty}^1 frac{mathrm d z}{1+z^2}\
          &=frac12left[arctan{z}right]_{-infty}^1
          end{align*}$$




          $$therefore~mathfrak I ~=~int_0^infty frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}mathrm d x~=~frac{3pi}8$$




          Where we used the symmetry aswell as well-known values of the tangent function.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 5 at 17:57

























          answered Jan 5 at 13:32









          mrtaurhomrtaurho

          5,70551540




          5,70551540























              6












              $begingroup$

              Well, we can just rewrite the integral as: $$int_0^inftyfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},dx=int_0^infty frac{x^2}{e^{2x}left(frac{1-x}{1+x}right)^2+e^{-2x}}frac{1}{(1+x)^2}dx$$
              $$=int_0^infty frac{1}{e^{4x}left(frac{1-x}{1+x}right)^2+1}frac{1}{e^{-2x}}frac{x^2}{(1+x)^2}dx =int_0^infty frac{color{blue}{e^{2x}}}{left(color{red}{e^{2x}cdotfrac{1-x}{1+x}}right)^2+1}color{blue}{frac{x^2}{(1+x)^2}dx}$$ Now set $,displaystyle{color{red}{e^{2x} frac{1-x}{1+x}}=tRightarrow color{blue}{e^{2x}frac{x^2}{(1+x)^2}dx}=-frac12 dt},$ and the result follows.






              share|cite|improve this answer











              $endgroup$


















                6












                $begingroup$

                Well, we can just rewrite the integral as: $$int_0^inftyfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},dx=int_0^infty frac{x^2}{e^{2x}left(frac{1-x}{1+x}right)^2+e^{-2x}}frac{1}{(1+x)^2}dx$$
                $$=int_0^infty frac{1}{e^{4x}left(frac{1-x}{1+x}right)^2+1}frac{1}{e^{-2x}}frac{x^2}{(1+x)^2}dx =int_0^infty frac{color{blue}{e^{2x}}}{left(color{red}{e^{2x}cdotfrac{1-x}{1+x}}right)^2+1}color{blue}{frac{x^2}{(1+x)^2}dx}$$ Now set $,displaystyle{color{red}{e^{2x} frac{1-x}{1+x}}=tRightarrow color{blue}{e^{2x}frac{x^2}{(1+x)^2}dx}=-frac12 dt},$ and the result follows.






                share|cite|improve this answer











                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  Well, we can just rewrite the integral as: $$int_0^inftyfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},dx=int_0^infty frac{x^2}{e^{2x}left(frac{1-x}{1+x}right)^2+e^{-2x}}frac{1}{(1+x)^2}dx$$
                  $$=int_0^infty frac{1}{e^{4x}left(frac{1-x}{1+x}right)^2+1}frac{1}{e^{-2x}}frac{x^2}{(1+x)^2}dx =int_0^infty frac{color{blue}{e^{2x}}}{left(color{red}{e^{2x}cdotfrac{1-x}{1+x}}right)^2+1}color{blue}{frac{x^2}{(1+x)^2}dx}$$ Now set $,displaystyle{color{red}{e^{2x} frac{1-x}{1+x}}=tRightarrow color{blue}{e^{2x}frac{x^2}{(1+x)^2}dx}=-frac12 dt},$ and the result follows.






                  share|cite|improve this answer











                  $endgroup$



                  Well, we can just rewrite the integral as: $$int_0^inftyfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},dx=int_0^infty frac{x^2}{e^{2x}left(frac{1-x}{1+x}right)^2+e^{-2x}}frac{1}{(1+x)^2}dx$$
                  $$=int_0^infty frac{1}{e^{4x}left(frac{1-x}{1+x}right)^2+1}frac{1}{e^{-2x}}frac{x^2}{(1+x)^2}dx =int_0^infty frac{color{blue}{e^{2x}}}{left(color{red}{e^{2x}cdotfrac{1-x}{1+x}}right)^2+1}color{blue}{frac{x^2}{(1+x)^2}dx}$$ Now set $,displaystyle{color{red}{e^{2x} frac{1-x}{1+x}}=tRightarrow color{blue}{e^{2x}frac{x^2}{(1+x)^2}dx}=-frac12 dt},$ and the result follows.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 5 at 13:03

























                  answered Jan 5 at 12:53









                  ZackyZacky

                  7,59511061




                  7,59511061























                      2












                      $begingroup$

                      The integral is $$int_0^xfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},mathrm dx=frac{1}{8} left(pi +4 tan ^{-1}left(frac{(x-1) e^{2x}}{x+1}right)right)$$ as can be verified by direct differentiation.



                      You can first tackle the $e^{2x}$ part by substitution. The rest should be a trivial matter.






                      share|cite|improve this answer









                      $endgroup$









                      • 4




                        $begingroup$
                        How did you do this?
                        $endgroup$
                        – Tyrell
                        Jan 5 at 12:44
















                      2












                      $begingroup$

                      The integral is $$int_0^xfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},mathrm dx=frac{1}{8} left(pi +4 tan ^{-1}left(frac{(x-1) e^{2x}}{x+1}right)right)$$ as can be verified by direct differentiation.



                      You can first tackle the $e^{2x}$ part by substitution. The rest should be a trivial matter.






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                      $endgroup$









                      • 4




                        $begingroup$
                        How did you do this?
                        $endgroup$
                        – Tyrell
                        Jan 5 at 12:44














                      2












                      2








                      2





                      $begingroup$

                      The integral is $$int_0^xfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},mathrm dx=frac{1}{8} left(pi +4 tan ^{-1}left(frac{(x-1) e^{2x}}{x+1}right)right)$$ as can be verified by direct differentiation.



                      You can first tackle the $e^{2x}$ part by substitution. The rest should be a trivial matter.






                      share|cite|improve this answer









                      $endgroup$



                      The integral is $$int_0^xfrac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2},mathrm dx=frac{1}{8} left(pi +4 tan ^{-1}left(frac{(x-1) e^{2x}}{x+1}right)right)$$ as can be verified by direct differentiation.



                      You can first tackle the $e^{2x}$ part by substitution. The rest should be a trivial matter.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 5 at 12:42









                      TreborTrebor

                      85415




                      85415








                      • 4




                        $begingroup$
                        How did you do this?
                        $endgroup$
                        – Tyrell
                        Jan 5 at 12:44














                      • 4




                        $begingroup$
                        How did you do this?
                        $endgroup$
                        – Tyrell
                        Jan 5 at 12:44








                      4




                      4




                      $begingroup$
                      How did you do this?
                      $endgroup$
                      – Tyrell
                      Jan 5 at 12:44




                      $begingroup$
                      How did you do this?
                      $endgroup$
                      – Tyrell
                      Jan 5 at 12:44



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