How can this integral be solved? Is is an indefinite integral?












1












$begingroup$



The integral I want to solve is $$int_1^2 frac {2 ln(x)}{x+1} dx$$




I tried to integrate it by parts in 2 ways and I tried to do the integral by parts twice, I thought of a change of variable and they all didn't work.



Can any one help by giving a solution please?






integration indefinite-integrals







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    1












    $begingroup$



    The integral I want to solve is $$int_1^2 frac {2 ln(x)}{x+1} dx$$




    I tried to integrate it by parts in 2 ways and I tried to do the integral by parts twice, I thought of a change of variable and they all didn't work.



    Can any one help by giving a solution please?






    integration indefinite-integrals







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      The integral I want to solve is $$int_1^2 frac {2 ln(x)}{x+1} dx$$




      I tried to integrate it by parts in 2 ways and I tried to do the integral by parts twice, I thought of a change of variable and they all didn't work.



      Can any one help by giving a solution please?






      integration indefinite-integrals







      share|cite|improve this question











      $endgroup$





      The integral I want to solve is $$int_1^2 frac {2 ln(x)}{x+1} dx$$




      I tried to integrate it by parts in 2 ways and I tried to do the integral by parts twice, I thought of a change of variable and they all didn't work.



      Can any one help by giving a solution please?






      integration indefinite-integrals




      integration indefinite-integrals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 23 '18 at 7:20









      Eevee Trainer

      6,1631936




      6,1631936










      asked Dec 23 '18 at 7:15









      Fareed AFFareed AF

      52612




      52612






















          1 Answer
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          $begingroup$

          Your integral evaluates to,
          $$2(ln (2)ln (3) + operatorname{Li}_2 (-2) + frac {π^2}{12})$$
          Use the identity,
          $$int (f.g') dx = f.g - int (f'.g) dx$$
          And the definition of dilogarithm to get your result.
          Assume $f(x):=ln x$ and $g'(x):=frac {1}{1+x}$ and substitute $z:=-x$.
          $$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @Fareed AF, the dilogarithm $Li_2$ is a special function like $log (x)$ and $e^x$ , it has its own series representation and identities. Read it here:en.m.wikipedia.org/wiki/Spence%27s_function
            $endgroup$
            – Awe Kumar Jha
            Dec 23 '18 at 9:32












          • $begingroup$
            Thank you :-), I appreciate your help
            $endgroup$
            – Fareed AF
            Dec 23 '18 at 11:05











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          $begingroup$

          Your integral evaluates to,
          $$2(ln (2)ln (3) + operatorname{Li}_2 (-2) + frac {π^2}{12})$$
          Use the identity,
          $$int (f.g') dx = f.g - int (f'.g) dx$$
          And the definition of dilogarithm to get your result.
          Assume $f(x):=ln x$ and $g'(x):=frac {1}{1+x}$ and substitute $z:=-x$.
          $$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @Fareed AF, the dilogarithm $Li_2$ is a special function like $log (x)$ and $e^x$ , it has its own series representation and identities. Read it here:en.m.wikipedia.org/wiki/Spence%27s_function
            $endgroup$
            – Awe Kumar Jha
            Dec 23 '18 at 9:32












          • $begingroup$
            Thank you :-), I appreciate your help
            $endgroup$
            – Fareed AF
            Dec 23 '18 at 11:05
















          2












          $begingroup$

          Your integral evaluates to,
          $$2(ln (2)ln (3) + operatorname{Li}_2 (-2) + frac {π^2}{12})$$
          Use the identity,
          $$int (f.g') dx = f.g - int (f'.g) dx$$
          And the definition of dilogarithm to get your result.
          Assume $f(x):=ln x$ and $g'(x):=frac {1}{1+x}$ and substitute $z:=-x$.
          $$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @Fareed AF, the dilogarithm $Li_2$ is a special function like $log (x)$ and $e^x$ , it has its own series representation and identities. Read it here:en.m.wikipedia.org/wiki/Spence%27s_function
            $endgroup$
            – Awe Kumar Jha
            Dec 23 '18 at 9:32












          • $begingroup$
            Thank you :-), I appreciate your help
            $endgroup$
            – Fareed AF
            Dec 23 '18 at 11:05














          2












          2








          2





          $begingroup$

          Your integral evaluates to,
          $$2(ln (2)ln (3) + operatorname{Li}_2 (-2) + frac {π^2}{12})$$
          Use the identity,
          $$int (f.g') dx = f.g - int (f'.g) dx$$
          And the definition of dilogarithm to get your result.
          Assume $f(x):=ln x$ and $g'(x):=frac {1}{1+x}$ and substitute $z:=-x$.
          $$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$






          share|cite|improve this answer











          $endgroup$



          Your integral evaluates to,
          $$2(ln (2)ln (3) + operatorname{Li}_2 (-2) + frac {π^2}{12})$$
          Use the identity,
          $$int (f.g') dx = f.g - int (f'.g) dx$$
          And the definition of dilogarithm to get your result.
          Assume $f(x):=ln x$ and $g'(x):=frac {1}{1+x}$ and substitute $z:=-x$.
          $$operatorname{Li}_2 := int frac {ln (1-t)}{t} dt$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 23 '18 at 7:40









          DavidG

          2,1321724




          2,1321724










          answered Dec 23 '18 at 7:25









          Awe Kumar JhaAwe Kumar Jha

          43813




          43813












          • $begingroup$
            @Fareed AF, the dilogarithm $Li_2$ is a special function like $log (x)$ and $e^x$ , it has its own series representation and identities. Read it here:en.m.wikipedia.org/wiki/Spence%27s_function
            $endgroup$
            – Awe Kumar Jha
            Dec 23 '18 at 9:32












          • $begingroup$
            Thank you :-), I appreciate your help
            $endgroup$
            – Fareed AF
            Dec 23 '18 at 11:05


















          • $begingroup$
            @Fareed AF, the dilogarithm $Li_2$ is a special function like $log (x)$ and $e^x$ , it has its own series representation and identities. Read it here:en.m.wikipedia.org/wiki/Spence%27s_function
            $endgroup$
            – Awe Kumar Jha
            Dec 23 '18 at 9:32












          • $begingroup$
            Thank you :-), I appreciate your help
            $endgroup$
            – Fareed AF
            Dec 23 '18 at 11:05
















          $begingroup$
          @Fareed AF, the dilogarithm $Li_2$ is a special function like $log (x)$ and $e^x$ , it has its own series representation and identities. Read it here:en.m.wikipedia.org/wiki/Spence%27s_function
          $endgroup$
          – Awe Kumar Jha
          Dec 23 '18 at 9:32






          $begingroup$
          @Fareed AF, the dilogarithm $Li_2$ is a special function like $log (x)$ and $e^x$ , it has its own series representation and identities. Read it here:en.m.wikipedia.org/wiki/Spence%27s_function
          $endgroup$
          – Awe Kumar Jha
          Dec 23 '18 at 9:32














          $begingroup$
          Thank you :-), I appreciate your help
          $endgroup$
          – Fareed AF
          Dec 23 '18 at 11:05




          $begingroup$
          Thank you :-), I appreciate your help
          $endgroup$
          – Fareed AF
          Dec 23 '18 at 11:05


















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