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Integral of $fracln(1+x^2)1+x^2$


How to find $int_0^inftyfracdx(1+x^2)^4}$Evaluating the integral $int_0^1arctan(1-x+x^2)dx$Closed-form expression of a definite integralHow do I find the following definite integral?How to integrate $frac{x^2log sin x1+x^6$Setting up this integral?Calculate $int_0^1fraccos(arctan x)sqrtxdx$Evaluate $intlimits_0^1fraclog(1-x+x^2)log(1+x-x^2)xdx$What is the simplest technique to evaluate the following definite triple integral?Indefinite integral $int arctan^2 x dx$ in terms of the dilogarithm function













1












$begingroup$


Compute $intfracln(1+x^2)1+x^2dx$



Attempt:



I tried to do Integration by Parts



$$u=ln(1+x^2)Rightarrow du=frac2xdx1+x^2\
dv=fracdx1+x^2Rightarrow v=arctan x\
intfracln(1+x^2)1+x^2dx=ln(1+x^2)arctan x-intfrac2xarctan x dx1+x^2$$



Since it seems that it not done in terms of elementary functions, what about the definite version:



$$int_0^1fracln(1+x^2)1+x^2dx$$






integration definite-integrals







share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    The solution containes the dilog-function
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 14 at 16:09










  • $begingroup$
    Could you possibly be interested in a definite version of this?
    $endgroup$
    – Math-fun
    Mar 14 at 16:11










  • $begingroup$
    so aparent the indefinite version inst doable in term of elementary functions? What about the definite integral with interval from $0$ to $1$ or like $r$ to $1$ where $0<r<1$.
    $endgroup$
    – cand
    Mar 14 at 16:15







  • 1




    $begingroup$
    That is what Maple have done:$$i/2ln left( -i/2 left( x+i right) right) ln left( x-i right) +i/4 left( ln left( x-i right) right) ^2-i/2ln left( x-i right) ln left( x^2+1 right) +i/2it dilog left( -i/2 left( x+i right) right) -i/4 left( ln left( x+i right) right) ^2-i/2ln left( x+i right) ln left( i/2 left( x-i right) right) +i/2ln left( x+i right) ln left( x^2+1 right) -i/2it dilog left( i/2 left( x-i right) right) $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 14 at 16:26















1












$begingroup$


Compute $intfracln(1+x^2)1+x^2dx$



Attempt:



I tried to do Integration by Parts



$$u=ln(1+x^2)Rightarrow du=frac2xdx1+x^2\
dv=fracdx1+x^2Rightarrow v=arctan x\
intfracln(1+x^2)1+x^2dx=ln(1+x^2)arctan x-intfrac2xarctan x dx1+x^2$$



Since it seems that it not done in terms of elementary functions, what about the definite version:



$$int_0^1fracln(1+x^2)1+x^2dx$$






integration definite-integrals







share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    The solution containes the dilog-function
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 14 at 16:09










  • $begingroup$
    Could you possibly be interested in a definite version of this?
    $endgroup$
    – Math-fun
    Mar 14 at 16:11










  • $begingroup$
    so aparent the indefinite version inst doable in term of elementary functions? What about the definite integral with interval from $0$ to $1$ or like $r$ to $1$ where $0<r<1$.
    $endgroup$
    – cand
    Mar 14 at 16:15







  • 1




    $begingroup$
    That is what Maple have done:$$i/2ln left( -i/2 left( x+i right) right) ln left( x-i right) +i/4 left( ln left( x-i right) right) ^2-i/2ln left( x-i right) ln left( x^2+1 right) +i/2it dilog left( -i/2 left( x+i right) right) -i/4 left( ln left( x+i right) right) ^2-i/2ln left( x+i right) ln left( i/2 left( x-i right) right) +i/2ln left( x+i right) ln left( x^2+1 right) -i/2it dilog left( i/2 left( x-i right) right) $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 14 at 16:26













1












1








1





$begingroup$


Compute $intfracln(1+x^2)1+x^2dx$



Attempt:



I tried to do Integration by Parts



$$u=ln(1+x^2)Rightarrow du=frac2xdx1+x^2\
dv=fracdx1+x^2Rightarrow v=arctan x\
intfracln(1+x^2)1+x^2dx=ln(1+x^2)arctan x-intfrac2xarctan x dx1+x^2$$



Since it seems that it not done in terms of elementary functions, what about the definite version:



$$int_0^1fracln(1+x^2)1+x^2dx$$






integration definite-integrals







share|cite|improve this question











$endgroup$




Compute $intfracln(1+x^2)1+x^2dx$



Attempt:



I tried to do Integration by Parts



$$u=ln(1+x^2)Rightarrow du=frac2xdx1+x^2\
dv=fracdx1+x^2Rightarrow v=arctan x\
intfracln(1+x^2)1+x^2dx=ln(1+x^2)arctan x-intfrac2xarctan x dx1+x^2$$



Since it seems that it not done in terms of elementary functions, what about the definite version:



$$int_0^1fracln(1+x^2)1+x^2dx$$






integration definite-integrals




integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 14 at 18:15









Paras Khosla

2,348222




2,348222










asked Mar 14 at 16:04









candcand

1,51911023




1,51911023







  • 2




    $begingroup$
    The solution containes the dilog-function
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 14 at 16:09










  • $begingroup$
    Could you possibly be interested in a definite version of this?
    $endgroup$
    – Math-fun
    Mar 14 at 16:11










  • $begingroup$
    so aparent the indefinite version inst doable in term of elementary functions? What about the definite integral with interval from $0$ to $1$ or like $r$ to $1$ where $0<r<1$.
    $endgroup$
    – cand
    Mar 14 at 16:15







  • 1




    $begingroup$
    That is what Maple have done:$$i/2ln left( -i/2 left( x+i right) right) ln left( x-i right) +i/4 left( ln left( x-i right) right) ^2-i/2ln left( x-i right) ln left( x^2+1 right) +i/2it dilog left( -i/2 left( x+i right) right) -i/4 left( ln left( x+i right) right) ^2-i/2ln left( x+i right) ln left( i/2 left( x-i right) right) +i/2ln left( x+i right) ln left( x^2+1 right) -i/2it dilog left( i/2 left( x-i right) right) $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 14 at 16:26












  • 2




    $begingroup$
    The solution containes the dilog-function
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 14 at 16:09










  • $begingroup$
    Could you possibly be interested in a definite version of this?
    $endgroup$
    – Math-fun
    Mar 14 at 16:11










  • $begingroup$
    so aparent the indefinite version inst doable in term of elementary functions? What about the definite integral with interval from $0$ to $1$ or like $r$ to $1$ where $0<r<1$.
    $endgroup$
    – cand
    Mar 14 at 16:15







  • 1




    $begingroup$
    That is what Maple have done:$$i/2ln left( -i/2 left( x+i right) right) ln left( x-i right) +i/4 left( ln left( x-i right) right) ^2-i/2ln left( x-i right) ln left( x^2+1 right) +i/2it dilog left( -i/2 left( x+i right) right) -i/4 left( ln left( x+i right) right) ^2-i/2ln left( x+i right) ln left( i/2 left( x-i right) right) +i/2ln left( x+i right) ln left( x^2+1 right) -i/2it dilog left( i/2 left( x-i right) right) $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 14 at 16:26







2




2




$begingroup$
The solution containes the dilog-function
$endgroup$
– Dr. Sonnhard Graubner
Mar 14 at 16:09




$begingroup$
The solution containes the dilog-function
$endgroup$
– Dr. Sonnhard Graubner
Mar 14 at 16:09












$begingroup$
Could you possibly be interested in a definite version of this?
$endgroup$
– Math-fun
Mar 14 at 16:11




$begingroup$
Could you possibly be interested in a definite version of this?
$endgroup$
– Math-fun
Mar 14 at 16:11












$begingroup$
so aparent the indefinite version inst doable in term of elementary functions? What about the definite integral with interval from $0$ to $1$ or like $r$ to $1$ where $0<r<1$.
$endgroup$
– cand
Mar 14 at 16:15





$begingroup$
so aparent the indefinite version inst doable in term of elementary functions? What about the definite integral with interval from $0$ to $1$ or like $r$ to $1$ where $0<r<1$.
$endgroup$
– cand
Mar 14 at 16:15





1




1




$begingroup$
That is what Maple have done:$$i/2ln left( -i/2 left( x+i right) right) ln left( x-i right) +i/4 left( ln left( x-i right) right) ^2-i/2ln left( x-i right) ln left( x^2+1 right) +i/2it dilog left( -i/2 left( x+i right) right) -i/4 left( ln left( x+i right) right) ^2-i/2ln left( x+i right) ln left( i/2 left( x-i right) right) +i/2ln left( x+i right) ln left( x^2+1 right) -i/2it dilog left( i/2 left( x-i right) right) $$
$endgroup$
– Dr. Sonnhard Graubner
Mar 14 at 16:26




$begingroup$
That is what Maple have done:$$i/2ln left( -i/2 left( x+i right) right) ln left( x-i right) +i/4 left( ln left( x-i right) right) ^2-i/2ln left( x-i right) ln left( x^2+1 right) +i/2it dilog left( -i/2 left( x+i right) right) -i/4 left( ln left( x+i right) right) ^2-i/2ln left( x+i right) ln left( i/2 left( x-i right) right) +i/2ln left( x+i right) ln left( x^2+1 right) -i/2it dilog left( i/2 left( x-i right) right) $$
$endgroup$
– Dr. Sonnhard Graubner
Mar 14 at 16:26










1 Answer
1






active

oldest

votes


















1












$begingroup$

$$textLet I=intdfracln(1+x^2)1+x^2mathrm dx=intleft(underbracedfraciln(1+x^2)2(x+i)_I_1-underbracedfraciln(1+x^2)2(x-i)_I_2right)mathrm dx$$



Solving $2/icdot I_1$:
$$beginbmatrixu \ mathrm duendbmatrix=beginbmatrixx+i\ mathrm dxendbmatrix$$ $$beginalignintdfracln(1+x^2)x+imathrm dx&=intdfracln((u-i)^2+1)umathrm du\ & =intdfracln(u-2i)umathrm du+dfrac12ln^2(u)\ & = intdfracln(iu/2+1)umathrm du+ln(-2i)ln u+dfrac12ln^2u \ &= -mathrmLi_2left(-dfraciu2right)+dfrac12ln^2u+ln(-2i)ln(u)\&=-mathrmLi_2left(-dfraci(x+i)2right)+dfrac12 ln^2(x+i)+ln(-2i)ln(x+i)tag1endalign$$



Similarly solve $2/icdot I_2$ to get the following result: $$intdfracln(1+x^2)x-imathrm dx = -mathrmLi_2left(-dfraci(x-i)2right)+dfrac12ln^2(x-i)+ln(2i)ln(x-i)tag2$$



Computing $i/2cdot [(1)+(2)]equiv I_1+I_2$ and simplifying gives you the required antiderivative stated as follows (drum-roll moment): $$I=dfraci4left[lnmid x+i mid left(lnmid x+imid+2ln(-2i)right)-lnmid x-imidleft(lnmid x-imid +2ln(2i)right)\ +2mathrmLi_2left(dfracix+12right)-2mathrmLi_2left(-dfracix-12right)right]+C$$






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    1






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    active

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    active

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    1












    $begingroup$

    $$textLet I=intdfracln(1+x^2)1+x^2mathrm dx=intleft(underbracedfraciln(1+x^2)2(x+i)_I_1-underbracedfraciln(1+x^2)2(x-i)_I_2right)mathrm dx$$



    Solving $2/icdot I_1$:
    $$beginbmatrixu \ mathrm duendbmatrix=beginbmatrixx+i\ mathrm dxendbmatrix$$ $$beginalignintdfracln(1+x^2)x+imathrm dx&=intdfracln((u-i)^2+1)umathrm du\ & =intdfracln(u-2i)umathrm du+dfrac12ln^2(u)\ & = intdfracln(iu/2+1)umathrm du+ln(-2i)ln u+dfrac12ln^2u \ &= -mathrmLi_2left(-dfraciu2right)+dfrac12ln^2u+ln(-2i)ln(u)\&=-mathrmLi_2left(-dfraci(x+i)2right)+dfrac12 ln^2(x+i)+ln(-2i)ln(x+i)tag1endalign$$



    Similarly solve $2/icdot I_2$ to get the following result: $$intdfracln(1+x^2)x-imathrm dx = -mathrmLi_2left(-dfraci(x-i)2right)+dfrac12ln^2(x-i)+ln(2i)ln(x-i)tag2$$



    Computing $i/2cdot [(1)+(2)]equiv I_1+I_2$ and simplifying gives you the required antiderivative stated as follows (drum-roll moment): $$I=dfraci4left[lnmid x+i mid left(lnmid x+imid+2ln(-2i)right)-lnmid x-imidleft(lnmid x-imid +2ln(2i)right)\ +2mathrmLi_2left(dfracix+12right)-2mathrmLi_2left(-dfracix-12right)right]+C$$






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      $$textLet I=intdfracln(1+x^2)1+x^2mathrm dx=intleft(underbracedfraciln(1+x^2)2(x+i)_I_1-underbracedfraciln(1+x^2)2(x-i)_I_2right)mathrm dx$$



      Solving $2/icdot I_1$:
      $$beginbmatrixu \ mathrm duendbmatrix=beginbmatrixx+i\ mathrm dxendbmatrix$$ $$beginalignintdfracln(1+x^2)x+imathrm dx&=intdfracln((u-i)^2+1)umathrm du\ & =intdfracln(u-2i)umathrm du+dfrac12ln^2(u)\ & = intdfracln(iu/2+1)umathrm du+ln(-2i)ln u+dfrac12ln^2u \ &= -mathrmLi_2left(-dfraciu2right)+dfrac12ln^2u+ln(-2i)ln(u)\&=-mathrmLi_2left(-dfraci(x+i)2right)+dfrac12 ln^2(x+i)+ln(-2i)ln(x+i)tag1endalign$$



      Similarly solve $2/icdot I_2$ to get the following result: $$intdfracln(1+x^2)x-imathrm dx = -mathrmLi_2left(-dfraci(x-i)2right)+dfrac12ln^2(x-i)+ln(2i)ln(x-i)tag2$$



      Computing $i/2cdot [(1)+(2)]equiv I_1+I_2$ and simplifying gives you the required antiderivative stated as follows (drum-roll moment): $$I=dfraci4left[lnmid x+i mid left(lnmid x+imid+2ln(-2i)right)-lnmid x-imidleft(lnmid x-imid +2ln(2i)right)\ +2mathrmLi_2left(dfracix+12right)-2mathrmLi_2left(-dfracix-12right)right]+C$$






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        $$textLet I=intdfracln(1+x^2)1+x^2mathrm dx=intleft(underbracedfraciln(1+x^2)2(x+i)_I_1-underbracedfraciln(1+x^2)2(x-i)_I_2right)mathrm dx$$



        Solving $2/icdot I_1$:
        $$beginbmatrixu \ mathrm duendbmatrix=beginbmatrixx+i\ mathrm dxendbmatrix$$ $$beginalignintdfracln(1+x^2)x+imathrm dx&=intdfracln((u-i)^2+1)umathrm du\ & =intdfracln(u-2i)umathrm du+dfrac12ln^2(u)\ & = intdfracln(iu/2+1)umathrm du+ln(-2i)ln u+dfrac12ln^2u \ &= -mathrmLi_2left(-dfraciu2right)+dfrac12ln^2u+ln(-2i)ln(u)\&=-mathrmLi_2left(-dfraci(x+i)2right)+dfrac12 ln^2(x+i)+ln(-2i)ln(x+i)tag1endalign$$



        Similarly solve $2/icdot I_2$ to get the following result: $$intdfracln(1+x^2)x-imathrm dx = -mathrmLi_2left(-dfraci(x-i)2right)+dfrac12ln^2(x-i)+ln(2i)ln(x-i)tag2$$



        Computing $i/2cdot [(1)+(2)]equiv I_1+I_2$ and simplifying gives you the required antiderivative stated as follows (drum-roll moment): $$I=dfraci4left[lnmid x+i mid left(lnmid x+imid+2ln(-2i)right)-lnmid x-imidleft(lnmid x-imid +2ln(2i)right)\ +2mathrmLi_2left(dfracix+12right)-2mathrmLi_2left(-dfracix-12right)right]+C$$






        share|cite|improve this answer









        $endgroup$



        $$textLet I=intdfracln(1+x^2)1+x^2mathrm dx=intleft(underbracedfraciln(1+x^2)2(x+i)_I_1-underbracedfraciln(1+x^2)2(x-i)_I_2right)mathrm dx$$



        Solving $2/icdot I_1$:
        $$beginbmatrixu \ mathrm duendbmatrix=beginbmatrixx+i\ mathrm dxendbmatrix$$ $$beginalignintdfracln(1+x^2)x+imathrm dx&=intdfracln((u-i)^2+1)umathrm du\ & =intdfracln(u-2i)umathrm du+dfrac12ln^2(u)\ & = intdfracln(iu/2+1)umathrm du+ln(-2i)ln u+dfrac12ln^2u \ &= -mathrmLi_2left(-dfraciu2right)+dfrac12ln^2u+ln(-2i)ln(u)\&=-mathrmLi_2left(-dfraci(x+i)2right)+dfrac12 ln^2(x+i)+ln(-2i)ln(x+i)tag1endalign$$



        Similarly solve $2/icdot I_2$ to get the following result: $$intdfracln(1+x^2)x-imathrm dx = -mathrmLi_2left(-dfraci(x-i)2right)+dfrac12ln^2(x-i)+ln(2i)ln(x-i)tag2$$



        Computing $i/2cdot [(1)+(2)]equiv I_1+I_2$ and simplifying gives you the required antiderivative stated as follows (drum-roll moment): $$I=dfraci4left[lnmid x+i mid left(lnmid x+imid+2ln(-2i)right)-lnmid x-imidleft(lnmid x-imid +2ln(2i)right)\ +2mathrmLi_2left(dfracix+12right)-2mathrmLi_2left(-dfracix-12right)right]+C$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 14 at 17:24









        Paras KhoslaParas Khosla

        2,348222




        2,348222



























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