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

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

Integre termo a termo:

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

Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=1$$$:

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

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

Então $$$du=\left(- x + e^{x}\right)^{\prime }dx = \left(e^{x} - 1\right) dx$$$ (veja os passos »), e obtemos $$$\left(e^{x} - 1\right) dx = du$$$.

A integral pode ser reescrita como

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

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

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

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

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

Portanto,

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

Adicione a constante de integração:

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

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