Integraal van $$$\frac{1}{e^{x} - 1}$$$

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

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

Dan $$$du=\left(e^{x}\right)^{\prime }dx = e^{x} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$e^{x} dx = du$$$.

Dus,

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

Voer een ontbinding in partiële breuken uit (stappen zijn te zien »):

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

Integreer termgewijs:

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

Zij $$$v=u - 1$$$.

Dan $$$dv=\left(u - 1\right)^{\prime }du = 1 du$$$ (de stappen zijn te zien »), en dan geldt dat $$$du = dv$$$.

Dus,

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

De integraal van $$$\frac{1}{v}$$$ is $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

We herinneren eraan dat $$$v=u - 1$$$:

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

De integraal van $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

We herinneren eraan dat $$$u=e^{x}$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

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