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![[tex]=ln(x)\frac{1}{2}ln(2x+1))-\int \frac{ln(2x+1)dx}{x}[/tex]](images/latex/b94fb465a0195bc60b85e95f861f7f203897fb45_0.png "[tex]=ln(x)\frac{1}{2}ln(2x+1))-\int \frac{ln(2x+1)dx}{x}[/tex]")
ahi me quedo porque cuando resuelvo la segunda integral vuelve a aparecerme un logaritmo.
![[tex]\int \frac{ln(x)}{2x+1}=ln(x)\frac{1}{2}ln(2x+1))-\int \frac{ln(2x+1)dx}{x}[/tex]](images/latex/04a632f2f79e6eaa6b1eead82eb953353ff1e6b4_0.png "[tex]\int \frac{ln(x)}{2x+1}=ln(x)\frac{1}{2}ln(2x+1))-\int \frac{ln(2x+1)dx}{x}[/tex]")
ahi me quedo porque cuando resuelvo la segunda integral vuelve a aparecerme un logaritmo.
![[tex] \int \frac{ln(x)}{2x+1}=ln(x)\frac{1}{2}ln(2x+1))-\int \frac{ln(2x+1)dx}{x} [/tex]](images/latex/2c31d5cb6439cc462b7acce780a3d338df10d766_0.png "[tex] \int \frac{ln(x)}{2x+1}=ln(x)\frac{1}{2}ln(2x+1))-\int \frac{ln(2x+1)dx}{x} [/tex]")
![[tex]u=ln(2x+1)[/tex]](images/latex/d6d29952792e5c0befb3c4b9cb4327932aff32fc_0.png "[tex]u=ln(2x+1)[/tex]")
![[tex]du=\frac{2}{2x+1}dx[/tex]](images/latex/6d837670ec3616de2e9a829d152c1adbee6e6644_0.png "[tex]du=\frac{2}{2x+1}dx[/tex]")
![[tex]2x=e^u-1[/tex]](images/latex/270c12bc260beda62bc13f75927d44774eb82ef3_0.png "[tex]2x=e^u-1[/tex]")
![[tex]\int \frac{u}{e^u-1}e^udu[/tex]](images/latex/666d9b65e9e83cd0fc9466402943adfadc34ebaa_0.png "[tex]\int \frac{u}{e^u-1}e^udu[/tex]")
![[tex]\frac{1}{2}ue^u-\frac{1}{2}\int e^udu= =\frac{1}{2}ue^u-\frac{1}{2}e^u[/tex]](images/latex/f60590935c231c9df0662918f7e4916047a0c4e1_0.png "[tex]\frac{1}{2}ue^u-\frac{1}{2}\int e^udu= =\frac{1}{2}ue^u-\frac{1}{2}e^u[/tex]")
![[tex]\int \frac{ln(x)}{2x+1}=ln(x)\frac{1}{2}ln(2x+1)-\int \frac{ln(2x+1)}{2x}=ln(x)\frac{1}{2}ln(2x+1)-\frac{1}{2}ue^u+\frac{1}{2}\int e^udu =ln(x)\frac{1}{2}ln(2x+1)-\frac{1}{2}ue^u+\frac{1}{2}e^u[/tex]](images/latex/98c327d58e06819fe71168372c4a998aae383262_0.png "[tex]\int \frac{ln(x)}{2x+1}=ln(x)\frac{1}{2}ln(2x+1)-\int \frac{ln(2x+1)}{2x}=ln(x)\frac{1}{2}ln(2x+1)-\frac{1}{2}ue^u+\frac{1}{2}\int e^udu =ln(x)\frac{1}{2}ln(2x+1)-\frac{1}{2}ue^u+\frac{1}{2}e^u[/tex]")
![[tex]\int \frac{ln(x)}{2x+1}=ln(x)\frac{1}{2}ln(2x+1)-\int \frac{ln(2x+1)}{2x}=ln(x)\frac{1}{2}ln(2x+1)-\frac{1}{2}ue^u+\frac{1}{2}\int e^udu =ln(x)\frac{1}{2}ln(2x+1)-\frac{1}{2}ln(2x+1)(2x+1)+\frac{1}{2}(2x+1)[/tex]](images/latex/47c8aaf1f59c2c9c5032fc13d74bbf018961727f_0.png "[tex]\int \frac{ln(x)}{2x+1}=ln(x)\frac{1}{2}ln(2x+1)-\int \frac{ln(2x+1)}{2x}=ln(x)\frac{1}{2}ln(2x+1)-\frac{1}{2}ue^u+\frac{1}{2}\int e^udu =ln(x)\frac{1}{2}ln(2x+1)-\frac{1}{2}ln(2x+1)(2x+1)+\frac{1}{2}(2x+1)[/tex]")
![[tex]\int \frac{e^u}{e^u-1}udu[/tex]](images/latex/528e8a310690488bbb6b42a84a5b02b8eff5a9a8_0.png "[tex]\int \frac{e^u}{e^u-1}udu[/tex]")
![[tex]\int \frac{e^u}{e^u-1}udu=ln(e^u-1)[/tex]](images/latex/806a2f7a7c311c67b6599d3f938683bb6c3da1bf_0.png "[tex]\int \frac{e^u}{e^u-1}udu=ln(e^u-1)[/tex]")
![[tex]\int \frac{e^u}{e^u-1}udu=ln(2x)[/tex]](images/latex/143df0e76d4b73c4d4853db3add5d4015f90e15a_0.png "[tex]\int \frac{e^u}{e^u-1}udu=ln(2x)[/tex]")
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