Integral de $$$-1 + \frac{e^{x} - 1}{- x + e^{x}}$$$

Halla $$$\int \left(-1 + \frac{e^{x} - 1}{- x + e^{x}}\right)\, dx$$$.

Integra término a término:

$${\color{red}{\int{\left(-1 + \frac{e^{x} - 1}{- x + e^{x}}\right)d x}}} = {\color{red}{\left(- \int{1 d x} + \int{\frac{e^{x} - 1}{- x + e^{x}} d x}\right)}}$$

Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=1$$$:

$$\int{\frac{e^{x} - 1}{- x + e^{x}} d x} - {\color{red}{\int{1 d x}}} = \int{\frac{e^{x} - 1}{- x + e^{x}} d x} - {\color{red}{x}}$$

Sea $$$u=- x + e^{x}$$$.

Entonces $$$du=\left(- x + e^{x}\right)^{\prime }dx = \left(e^{x} - 1\right) dx$$$ (los pasos pueden verse »), y obtenemos que $$$\left(e^{x} - 1\right) dx = du$$$.

La integral puede reescribirse como

$$- x + {\color{red}{\int{\frac{e^{x} - 1}{- x + e^{x}} d x}}} = - x + {\color{red}{\int{\frac{1}{u} d u}}}$$

La integral de $$$\frac{1}{u}$$$ es $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$- x + {\color{red}{\int{\frac{1}{u} d u}}} = - x + {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Recordemos que $$$u=- x + e^{x}$$$:

$$- x + \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = - x + \ln{\left(\left|{{\color{red}{\left(- x + e^{x}\right)}}}\right| \right)}$$

Por lo tanto,

$$\int{\left(-1 + \frac{e^{x} - 1}{- x + e^{x}}\right)d x} = - x + \ln{\left(\left|{x - e^{x}}\right| \right)}$$

Añade la constante de integración:

$$\int{\left(-1 + \frac{e^{x} - 1}{- x + e^{x}}\right)d x} = - x + \ln{\left(\left|{x - e^{x}}\right| \right)}+C$$

$$$\int \left(-1 + \frac{e^{x} - 1}{- x + e^{x}}\right)\, dx = \left(- x + \ln\left(\left|{x - e^{x}}\right|\right)\right) + C$$$A