Evaluating the integral $int_0^1 fracln(1+x^2)arctan(x^2)dx1+x^2$

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$$int_0^1 fracln(1+x^2)arctan(x^2)dx1+x^2$$



I couldn't think of much except taking $x=tan(t)$ which reduced the integral to $$int_0^pi/4ln(sec^2(t))arctan(tan^2(t))dt$$



Which still doesn't looks good to me.



Taking $arctan(x^2)=t$ doesn't help much either. Neither does $ln(1+x^2)=t$




integration definite-integrals




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  • Why do you think this has a nice answer? Where did you find it?
    – Frpzzd
    Jul 14 at 20:41










  • To be honest, my friend gave me this. Firstly I spent time on this myself, then decided to post here. It might aswell be true that it doesn't have a nice answer but Wolfram returned a value of 0.8
    – So Lo
    Jul 14 at 20:54











  • $arctan(tan^2(t))=-arctan(cos(2t))+fracpi4$ But that probably doesn't help... I think your "friend" made a mistake, this is no "fun" question to do in your sparetime ;-)
    – GambitSquared
    Jul 14 at 20:55











  • have you tried uv method?
    – Nick
    Jul 14 at 23:28






  • 3




    Are you sure that the problem is not $int_0^1 fracln(1+x^2)arctan(x)dx1+x^2$ instead ?
    – Claude Leibovici
    Jul 15 at 5:33














up vote
1
down vote

favorite
1












$$int_0^1 fracln(1+x^2)arctan(x^2)dx1+x^2$$



I couldn't think of much except taking $x=tan(t)$ which reduced the integral to $$int_0^pi/4ln(sec^2(t))arctan(tan^2(t))dt$$



Which still doesn't looks good to me.



Taking $arctan(x^2)=t$ doesn't help much either. Neither does $ln(1+x^2)=t$




integration definite-integrals




share|cite|improve this question





















  • Why do you think this has a nice answer? Where did you find it?
    – Frpzzd
    Jul 14 at 20:41










  • To be honest, my friend gave me this. Firstly I spent time on this myself, then decided to post here. It might aswell be true that it doesn't have a nice answer but Wolfram returned a value of 0.8
    – So Lo
    Jul 14 at 20:54











  • $arctan(tan^2(t))=-arctan(cos(2t))+fracpi4$ But that probably doesn't help... I think your "friend" made a mistake, this is no "fun" question to do in your sparetime ;-)
    – GambitSquared
    Jul 14 at 20:55











  • have you tried uv method?
    – Nick
    Jul 14 at 23:28






  • 3




    Are you sure that the problem is not $int_0^1 fracln(1+x^2)arctan(x)dx1+x^2$ instead ?
    – Claude Leibovici
    Jul 15 at 5:33












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





$$int_0^1 fracln(1+x^2)arctan(x^2)dx1+x^2$$



I couldn't think of much except taking $x=tan(t)$ which reduced the integral to $$int_0^pi/4ln(sec^2(t))arctan(tan^2(t))dt$$



Which still doesn't looks good to me.



Taking $arctan(x^2)=t$ doesn't help much either. Neither does $ln(1+x^2)=t$




integration definite-integrals




share|cite|improve this question













$$int_0^1 fracln(1+x^2)arctan(x^2)dx1+x^2$$



I couldn't think of much except taking $x=tan(t)$ which reduced the integral to $$int_0^pi/4ln(sec^2(t))arctan(tan^2(t))dt$$



Which still doesn't looks good to me.



Taking $arctan(x^2)=t$ doesn't help much either. Neither does $ln(1+x^2)=t$





integration definite-integrals





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 14 at 19:38









José Carlos Santos

114k1698177




114k1698177









asked Jul 14 at 19:22









So Lo

5168




5168











  • Why do you think this has a nice answer? Where did you find it?
    – Frpzzd
    Jul 14 at 20:41










  • To be honest, my friend gave me this. Firstly I spent time on this myself, then decided to post here. It might aswell be true that it doesn't have a nice answer but Wolfram returned a value of 0.8
    – So Lo
    Jul 14 at 20:54











  • $arctan(tan^2(t))=-arctan(cos(2t))+fracpi4$ But that probably doesn't help... I think your "friend" made a mistake, this is no "fun" question to do in your sparetime ;-)
    – GambitSquared
    Jul 14 at 20:55











  • have you tried uv method?
    – Nick
    Jul 14 at 23:28






  • 3




    Are you sure that the problem is not $int_0^1 fracln(1+x^2)arctan(x)dx1+x^2$ instead ?
    – Claude Leibovici
    Jul 15 at 5:33
















  • Why do you think this has a nice answer? Where did you find it?
    – Frpzzd
    Jul 14 at 20:41










  • To be honest, my friend gave me this. Firstly I spent time on this myself, then decided to post here. It might aswell be true that it doesn't have a nice answer but Wolfram returned a value of 0.8
    – So Lo
    Jul 14 at 20:54











  • $arctan(tan^2(t))=-arctan(cos(2t))+fracpi4$ But that probably doesn't help... I think your "friend" made a mistake, this is no "fun" question to do in your sparetime ;-)
    – GambitSquared
    Jul 14 at 20:55











  • have you tried uv method?
    – Nick
    Jul 14 at 23:28






  • 3




    Are you sure that the problem is not $int_0^1 fracln(1+x^2)arctan(x)dx1+x^2$ instead ?
    – Claude Leibovici
    Jul 15 at 5:33















Why do you think this has a nice answer? Where did you find it?
– Frpzzd
Jul 14 at 20:41




Why do you think this has a nice answer? Where did you find it?
– Frpzzd
Jul 14 at 20:41












To be honest, my friend gave me this. Firstly I spent time on this myself, then decided to post here. It might aswell be true that it doesn't have a nice answer but Wolfram returned a value of 0.8
– So Lo
Jul 14 at 20:54





To be honest, my friend gave me this. Firstly I spent time on this myself, then decided to post here. It might aswell be true that it doesn't have a nice answer but Wolfram returned a value of 0.8
– So Lo
Jul 14 at 20:54













$arctan(tan^2(t))=-arctan(cos(2t))+fracpi4$ But that probably doesn't help... I think your "friend" made a mistake, this is no "fun" question to do in your sparetime ;-)
– GambitSquared
Jul 14 at 20:55





$arctan(tan^2(t))=-arctan(cos(2t))+fracpi4$ But that probably doesn't help... I think your "friend" made a mistake, this is no "fun" question to do in your sparetime ;-)
– GambitSquared
Jul 14 at 20:55













have you tried uv method?
– Nick
Jul 14 at 23:28




have you tried uv method?
– Nick
Jul 14 at 23:28




3




3




Are you sure that the problem is not $int_0^1 fracln(1+x^2)arctan(x)dx1+x^2$ instead ?
– Claude Leibovici
Jul 15 at 5:33




Are you sure that the problem is not $int_0^1 fracln(1+x^2)arctan(x)dx1+x^2$ instead ?
– Claude Leibovici
Jul 15 at 5:33















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