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

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

Integralkan suku demi suku:

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

Terapkan aturan konstanta $$$\int c\, dx = c x$$$ dengan $$$c=1$$$:

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

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

Kemudian $$$du=\left(- x + e^{x}\right)^{\prime }dx = \left(e^{x} - 1\right) dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$\left(e^{x} - 1\right) dx = du$$$.

Integralnya menjadi

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

Integral dari $$$\frac{1}{u}$$$ adalah $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Ingat bahwa $$$u=- x + e^{x}$$$:

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

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

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