How to solve $intfrac{ln x}{x^2(ln (x)-1)^2}dx$ by substitution











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$$intfrac{ln x}{x^2(ln (x)-1)^2}dx$$



Hello, I haven't be able to solve this integral, I've tried to do $u = ln (x)-1$ but couldn't make it work, any insight?










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    up vote
    6
    down vote

    favorite












    $$intfrac{ln x}{x^2(ln (x)-1)^2}dx$$



    Hello, I haven't be able to solve this integral, I've tried to do $u = ln (x)-1$ but couldn't make it work, any insight?










    share|cite|improve this question


























      up vote
      6
      down vote

      favorite









      up vote
      6
      down vote

      favorite











      $$intfrac{ln x}{x^2(ln (x)-1)^2}dx$$



      Hello, I haven't be able to solve this integral, I've tried to do $u = ln (x)-1$ but couldn't make it work, any insight?










      share|cite|improve this question















      $$intfrac{ln x}{x^2(ln (x)-1)^2}dx$$



      Hello, I haven't be able to solve this integral, I've tried to do $u = ln (x)-1$ but couldn't make it work, any insight?







      calculus integration indefinite-integrals






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      edited Nov 24 at 6:15









      user21820

      38.1k541150




      38.1k541150










      asked Nov 23 at 21:05









      Andres Oropeza

      355




      355






















          1 Answer
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          $xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$






          share|cite|improve this answer

















          • 5




            I think it should be noted that the motivation of the substitution is because the denominator can be rewritten as $(x(ln x-1))^2$, and of course the nice fact that $du/dx$ turns out exactly to be the numerator.
            – YiFan
            Nov 23 at 21:39











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          16
          down vote



          accepted










          $xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$






          share|cite|improve this answer

















          • 5




            I think it should be noted that the motivation of the substitution is because the denominator can be rewritten as $(x(ln x-1))^2$, and of course the nice fact that $du/dx$ turns out exactly to be the numerator.
            – YiFan
            Nov 23 at 21:39















          up vote
          16
          down vote



          accepted










          $xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$






          share|cite|improve this answer

















          • 5




            I think it should be noted that the motivation of the substitution is because the denominator can be rewritten as $(x(ln x-1))^2$, and of course the nice fact that $du/dx$ turns out exactly to be the numerator.
            – YiFan
            Nov 23 at 21:39













          up vote
          16
          down vote



          accepted







          up vote
          16
          down vote



          accepted






          $xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$






          share|cite|improve this answer












          $xln x-x=uimplies ln x dx=duimpliesdisplaystyle intfrac{du}{u^2}=dfrac{u^{-1}}{-1}+C=dfrac{1}{x-xln x}+C.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 at 21:20









          Yadati Kiran

          1,243417




          1,243417








          • 5




            I think it should be noted that the motivation of the substitution is because the denominator can be rewritten as $(x(ln x-1))^2$, and of course the nice fact that $du/dx$ turns out exactly to be the numerator.
            – YiFan
            Nov 23 at 21:39














          • 5




            I think it should be noted that the motivation of the substitution is because the denominator can be rewritten as $(x(ln x-1))^2$, and of course the nice fact that $du/dx$ turns out exactly to be the numerator.
            – YiFan
            Nov 23 at 21:39








          5




          5




          I think it should be noted that the motivation of the substitution is because the denominator can be rewritten as $(x(ln x-1))^2$, and of course the nice fact that $du/dx$ turns out exactly to be the numerator.
          – YiFan
          Nov 23 at 21:39




          I think it should be noted that the motivation of the substitution is because the denominator can be rewritten as $(x(ln x-1))^2$, and of course the nice fact that $du/dx$ turns out exactly to be the numerator.
          – YiFan
          Nov 23 at 21:39


















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