Evaluate $int{frac{e^{2x}} {sqrt{1-e^x}}} dx$












3














Evaluate $$displaystyleint{frac{e^{2x}} {sqrt{1-e^x}}} dx.$$



I tried to solve by using integration by parts, but I couldn't find a solution. What method should I use to integrate this?






integration indefinite-integrals







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  • Welcome to Stack Exchange. Can you tell us what you tried to do to solve this problem?
    – Asimov
    May 19 '14 at 1:39






  • 1




    Is that supposed to be $e^{2x}$ or $e^2x$, as this drastically changes the difficulty of the integral
    – Triatticus
    May 19 '14 at 1:45


















3














Evaluate $$displaystyleint{frac{e^{2x}} {sqrt{1-e^x}}} dx.$$



I tried to solve by using integration by parts, but I couldn't find a solution. What method should I use to integrate this?






integration indefinite-integrals







share|cite|improve this question
























  • Welcome to Stack Exchange. Can you tell us what you tried to do to solve this problem?
    – Asimov
    May 19 '14 at 1:39






  • 1




    Is that supposed to be $e^{2x}$ or $e^2x$, as this drastically changes the difficulty of the integral
    – Triatticus
    May 19 '14 at 1:45
















3












3








3







Evaluate $$displaystyleint{frac{e^{2x}} {sqrt{1-e^x}}} dx.$$



I tried to solve by using integration by parts, but I couldn't find a solution. What method should I use to integrate this?






integration indefinite-integrals







share|cite|improve this question















Evaluate $$displaystyleint{frac{e^{2x}} {sqrt{1-e^x}}} dx.$$



I tried to solve by using integration by parts, but I couldn't find a solution. What method should I use to integrate this?






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 Nov 24 at 6:17









Martin Sleziak

44.6k7115270




44.6k7115270










asked May 19 '14 at 1:24









user151911

162




162












  • Welcome to Stack Exchange. Can you tell us what you tried to do to solve this problem?
    – Asimov
    May 19 '14 at 1:39






  • 1




    Is that supposed to be $e^{2x}$ or $e^2x$, as this drastically changes the difficulty of the integral
    – Triatticus
    May 19 '14 at 1:45




















  • Welcome to Stack Exchange. Can you tell us what you tried to do to solve this problem?
    – Asimov
    May 19 '14 at 1:39






  • 1




    Is that supposed to be $e^{2x}$ or $e^2x$, as this drastically changes the difficulty of the integral
    – Triatticus
    May 19 '14 at 1:45


















Welcome to Stack Exchange. Can you tell us what you tried to do to solve this problem?
– Asimov
May 19 '14 at 1:39




Welcome to Stack Exchange. Can you tell us what you tried to do to solve this problem?
– Asimov
May 19 '14 at 1:39




1




1




Is that supposed to be $e^{2x}$ or $e^2x$, as this drastically changes the difficulty of the integral
– Triatticus
May 19 '14 at 1:45






Is that supposed to be $e^{2x}$ or $e^2x$, as this drastically changes the difficulty of the integral
– Triatticus
May 19 '14 at 1:45












3 Answers
3






active

oldest

votes


















6














Hint: Let $u=1-e^{x}$${}{}{}{}{}{}{}$






share|cite|improve this answer





















  • by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2
    – user151911
    May 19 '14 at 1:43






  • 1




    An alternative way to solve this using integration by parts is to notice that $frac{e^{2x}}{sqrt{1-e^{x}}}=frac{e^{x}}{sqrt{1-e^{x}}}e^{x}=bigg(frac{d}{dx}(2 sqrt{1-e^{x}})bigg)e^{x}$
    – user71352
    May 19 '14 at 1:46





















0














$$
intfrac{e^x}{sqrt{1-e^x}}underbrace{Big(e^x , dxBig)}_{text{HINT}}
$$



After that substitution (the one that is hinted at), use another substitution, namely a rationalizing substitution.






share|cite|improve this answer





























    0














    $$I=intfrac{e^{2x}}{(1-e^x)^{1/2}}dx$$
    $$I=intfrac{e^x}{(1-e^x)^{1/2}}e^xdx$$
    Substitution: $u^2=1-e^xRightarrow -2udu=e^xdx$.
    $$I=intfrac{1-u^2}{(u^2)^{1/2}}(-2)udu$$
    $$I=-2int(1-u^2)du$$
    $$I=2int(u^2-1)du$$
    $$I=2int u^2du-2int du$$
    $$I=frac{2u^4}{4}-2u$$
    $$I=frac{(1-e^x)^2}{2}-2sqrt{1-e^x}+C$$






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6














      Hint: Let $u=1-e^{x}$${}{}{}{}{}{}{}$






      share|cite|improve this answer





















      • by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2
        – user151911
        May 19 '14 at 1:43






      • 1




        An alternative way to solve this using integration by parts is to notice that $frac{e^{2x}}{sqrt{1-e^{x}}}=frac{e^{x}}{sqrt{1-e^{x}}}e^{x}=bigg(frac{d}{dx}(2 sqrt{1-e^{x}})bigg)e^{x}$
        – user71352
        May 19 '14 at 1:46


















      6














      Hint: Let $u=1-e^{x}$${}{}{}{}{}{}{}$






      share|cite|improve this answer





















      • by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2
        – user151911
        May 19 '14 at 1:43






      • 1




        An alternative way to solve this using integration by parts is to notice that $frac{e^{2x}}{sqrt{1-e^{x}}}=frac{e^{x}}{sqrt{1-e^{x}}}e^{x}=bigg(frac{d}{dx}(2 sqrt{1-e^{x}})bigg)e^{x}$
        – user71352
        May 19 '14 at 1:46
















      6












      6








      6






      Hint: Let $u=1-e^{x}$${}{}{}{}{}{}{}$






      share|cite|improve this answer












      Hint: Let $u=1-e^{x}$${}{}{}{}{}{}{}$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered May 19 '14 at 1:38









      user71352

      11.4k21025




      11.4k21025












      • by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2
        – user151911
        May 19 '14 at 1:43






      • 1




        An alternative way to solve this using integration by parts is to notice that $frac{e^{2x}}{sqrt{1-e^{x}}}=frac{e^{x}}{sqrt{1-e^{x}}}e^{x}=bigg(frac{d}{dx}(2 sqrt{1-e^{x}})bigg)e^{x}$
        – user71352
        May 19 '14 at 1:46




















      • by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2
        – user151911
        May 19 '14 at 1:43






      • 1




        An alternative way to solve this using integration by parts is to notice that $frac{e^{2x}}{sqrt{1-e^{x}}}=frac{e^{x}}{sqrt{1-e^{x}}}e^{x}=bigg(frac{d}{dx}(2 sqrt{1-e^{x}})bigg)e^{x}$
        – user71352
        May 19 '14 at 1:46


















      by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2
      – user151911
      May 19 '14 at 1:43




      by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2
      – user151911
      May 19 '14 at 1:43




      1




      1




      An alternative way to solve this using integration by parts is to notice that $frac{e^{2x}}{sqrt{1-e^{x}}}=frac{e^{x}}{sqrt{1-e^{x}}}e^{x}=bigg(frac{d}{dx}(2 sqrt{1-e^{x}})bigg)e^{x}$
      – user71352
      May 19 '14 at 1:46






      An alternative way to solve this using integration by parts is to notice that $frac{e^{2x}}{sqrt{1-e^{x}}}=frac{e^{x}}{sqrt{1-e^{x}}}e^{x}=bigg(frac{d}{dx}(2 sqrt{1-e^{x}})bigg)e^{x}$
      – user71352
      May 19 '14 at 1:46













      0














      $$
      intfrac{e^x}{sqrt{1-e^x}}underbrace{Big(e^x , dxBig)}_{text{HINT}}
      $$



      After that substitution (the one that is hinted at), use another substitution, namely a rationalizing substitution.






      share|cite|improve this answer


























        0














        $$
        intfrac{e^x}{sqrt{1-e^x}}underbrace{Big(e^x , dxBig)}_{text{HINT}}
        $$



        After that substitution (the one that is hinted at), use another substitution, namely a rationalizing substitution.






        share|cite|improve this answer
























          0












          0








          0






          $$
          intfrac{e^x}{sqrt{1-e^x}}underbrace{Big(e^x , dxBig)}_{text{HINT}}
          $$



          After that substitution (the one that is hinted at), use another substitution, namely a rationalizing substitution.






          share|cite|improve this answer












          $$
          intfrac{e^x}{sqrt{1-e^x}}underbrace{Big(e^x , dxBig)}_{text{HINT}}
          $$



          After that substitution (the one that is hinted at), use another substitution, namely a rationalizing substitution.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 19 '14 at 2:52









          Michael Hardy

          1




          1























              0














              $$I=intfrac{e^{2x}}{(1-e^x)^{1/2}}dx$$
              $$I=intfrac{e^x}{(1-e^x)^{1/2}}e^xdx$$
              Substitution: $u^2=1-e^xRightarrow -2udu=e^xdx$.
              $$I=intfrac{1-u^2}{(u^2)^{1/2}}(-2)udu$$
              $$I=-2int(1-u^2)du$$
              $$I=2int(u^2-1)du$$
              $$I=2int u^2du-2int du$$
              $$I=frac{2u^4}{4}-2u$$
              $$I=frac{(1-e^x)^2}{2}-2sqrt{1-e^x}+C$$






              share|cite|improve this answer


























                0














                $$I=intfrac{e^{2x}}{(1-e^x)^{1/2}}dx$$
                $$I=intfrac{e^x}{(1-e^x)^{1/2}}e^xdx$$
                Substitution: $u^2=1-e^xRightarrow -2udu=e^xdx$.
                $$I=intfrac{1-u^2}{(u^2)^{1/2}}(-2)udu$$
                $$I=-2int(1-u^2)du$$
                $$I=2int(u^2-1)du$$
                $$I=2int u^2du-2int du$$
                $$I=frac{2u^4}{4}-2u$$
                $$I=frac{(1-e^x)^2}{2}-2sqrt{1-e^x}+C$$






                share|cite|improve this answer
























                  0












                  0








                  0






                  $$I=intfrac{e^{2x}}{(1-e^x)^{1/2}}dx$$
                  $$I=intfrac{e^x}{(1-e^x)^{1/2}}e^xdx$$
                  Substitution: $u^2=1-e^xRightarrow -2udu=e^xdx$.
                  $$I=intfrac{1-u^2}{(u^2)^{1/2}}(-2)udu$$
                  $$I=-2int(1-u^2)du$$
                  $$I=2int(u^2-1)du$$
                  $$I=2int u^2du-2int du$$
                  $$I=frac{2u^4}{4}-2u$$
                  $$I=frac{(1-e^x)^2}{2}-2sqrt{1-e^x}+C$$






                  share|cite|improve this answer












                  $$I=intfrac{e^{2x}}{(1-e^x)^{1/2}}dx$$
                  $$I=intfrac{e^x}{(1-e^x)^{1/2}}e^xdx$$
                  Substitution: $u^2=1-e^xRightarrow -2udu=e^xdx$.
                  $$I=intfrac{1-u^2}{(u^2)^{1/2}}(-2)udu$$
                  $$I=-2int(1-u^2)du$$
                  $$I=2int(u^2-1)du$$
                  $$I=2int u^2du-2int du$$
                  $$I=frac{2u^4}{4}-2u$$
                  $$I=frac{(1-e^x)^2}{2}-2sqrt{1-e^x}+C$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 11 at 21:22









                  clathratus

                  3,022330




                  3,022330






























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