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

Bepaal $$$\int \left(- \frac{1}{1 - x}\right)\, dx$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=-1$$$ en $$$f{\left(x \right)} = \frac{1}{1 - x}$$$:

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

Zij $$$u=1 - x$$$.

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

De integraal kan worden herschreven als

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

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=-1$$$ en $$$f{\left(u \right)} = \frac{1}{u}$$$:

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

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

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

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

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

Dus,

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

Voeg de integratieconstante toe:

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

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