Integraal van $$$\ln\left(\frac{1}{1 - x}\right)$$$

Bepaal $$$\int \left(- \ln\left(1 - x\right)\right)\, dx$$$.

De invoer is herschreven: $$$\int{\ln{\left(\frac{1}{1 - x} \right)} d x}=\int{\left(- \ln{\left(1 - x \right)}\right)d x}$$$.

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

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

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

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

Dus,

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

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

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

Voor de integraal $$$\int{\ln{\left(u \right)} d u}$$$, gebruik partiële integratie $$$\int \operatorname{\kappa} \operatorname{dv} = \operatorname{\kappa}\operatorname{v} - \int \operatorname{v} \operatorname{d\kappa}$$$.

Zij $$$\operatorname{\kappa}=\ln{\left(u \right)}$$$ en $$$\operatorname{dv}=du$$$.

Dan $$$\operatorname{d\kappa}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (de stappen zijn te zien ») en $$$\operatorname{v}=\int{1 d u}=u$$$ (de stappen zijn te zien »).

Dus,

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

Pas de constantenregel $$$\int c\, du = c u$$$ toe met $$$c=1$$$:

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

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

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

Dus,

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

Vereenvoudig:

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

Voeg de constante van integratie toe (en verwijder de constante uit de uitdrukking):

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

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