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Integral de $$$\frac{1}{e^{x} - 1}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \frac{1}{e^{x} - 1}\, dx$$$.
Solución
Sea $$$u=e^{x}$$$.
Entonces $$$du=\left(e^{x}\right)^{\prime }dx = e^{x} dx$$$ (los pasos pueden verse »), y obtenemos que $$$e^{x} dx = du$$$.
La integral puede reescribirse como
$${\color{red}{\int{\frac{1}{e^{x} - 1} d x}}} = {\color{red}{\int{\frac{1}{u \left(u - 1\right)} d u}}}$$
Realizar la descomposición en fracciones parciales (los pasos pueden verse »):
$${\color{red}{\int{\frac{1}{u \left(u - 1\right)} d u}}} = {\color{red}{\int{\left(\frac{1}{u - 1} - \frac{1}{u}\right)d u}}}$$
Integra término a término:
$${\color{red}{\int{\left(\frac{1}{u - 1} - \frac{1}{u}\right)d u}}} = {\color{red}{\left(- \int{\frac{1}{u} d u} + \int{\frac{1}{u - 1} d u}\right)}}$$
Sea $$$v=u - 1$$$.
Entonces $$$dv=\left(u - 1\right)^{\prime }du = 1 du$$$ (los pasos pueden verse »), y obtenemos que $$$du = dv$$$.
Por lo tanto,
$$- \int{\frac{1}{u} d u} + {\color{red}{\int{\frac{1}{u - 1} d u}}} = - \int{\frac{1}{u} d u} + {\color{red}{\int{\frac{1}{v} d v}}}$$
La integral de $$$\frac{1}{v}$$$ es $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:
$$- \int{\frac{1}{u} d u} + {\color{red}{\int{\frac{1}{v} d v}}} = - \int{\frac{1}{u} d u} + {\color{red}{\ln{\left(\left|{v}\right| \right)}}}$$
Recordemos que $$$v=u - 1$$$:
$$\ln{\left(\left|{{\color{red}{v}}}\right| \right)} - \int{\frac{1}{u} d u} = \ln{\left(\left|{{\color{red}{\left(u - 1\right)}}}\right| \right)} - \int{\frac{1}{u} d u}$$
La integral de $$$\frac{1}{u}$$$ es $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$\ln{\left(\left|{u - 1}\right| \right)} - {\color{red}{\int{\frac{1}{u} d u}}} = \ln{\left(\left|{u - 1}\right| \right)} - {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
Recordemos que $$$u=e^{x}$$$:
$$\ln{\left(\left|{-1 + {\color{red}{u}}}\right| \right)} - \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{-1 + {\color{red}{e^{x}}}}\right| \right)} - \ln{\left(\left|{{\color{red}{e^{x}}}}\right| \right)}$$
Por lo tanto,
$$\int{\frac{1}{e^{x} - 1} d x} = - x + \ln{\left(\left|{e^{x} - 1}\right| \right)}$$
Añade la constante de integración:
$$\int{\frac{1}{e^{x} - 1} d x} = - x + \ln{\left(\left|{e^{x} - 1}\right| \right)}+C$$
Respuesta
$$$\int \frac{1}{e^{x} - 1}\, dx = \left(- x + \ln\left(\left|{e^{x} - 1}\right|\right)\right) + C$$$A