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

Temukan $$$\int \frac{1}{e^{x} - 1}\, dx$$$.

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

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

Integralnya menjadi

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

Lakukan dekomposisi pecahan parsial (langkah-langkah dapat dilihat di »):

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

Integralkan suku demi suku:

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

Misalkan $$$v=u - 1$$$.

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

Jadi,

$$- \int{\frac{1}{u} d u} + {\color{red}{\int{\frac{1}{u - 1} d u}}} = - \int{\frac{1}{u} d u} + {\color{red}{\int{\frac{1}{v} d v}}}$$

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

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

Ingat bahwa $$$v=u - 1$$$:

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

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

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

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

$$\ln{\left(\left|{-1 + {\color{red}{u}}}\right| \right)} - \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{-1 + {\color{red}{e^{x}}}}\right| \right)} - \ln{\left(\left|{{\color{red}{e^{x}}}}\right| \right)}$$

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

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