Integraal van $$$\ln\left(x - 5\right)$$$

Bepaal $$$\int \ln\left(x - 5\right)\, dx$$$.

Zij $$$u=x - 5$$$.

Dan $$$du=\left(x - 5\right)^{\prime }dx = 1 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = du$$$.

Dus,

$${\color{red}{\int{\ln{\left(x - 5 \right)} d x}}} = {\color{red}{\int{\ln{\left(u \right)} d u}}}$$

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

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

Dan $$$\operatorname{dc}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (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 $$$u=x - 5$$$:

$$- {\color{red}{u}} + {\color{red}{u}} \ln{\left({\color{red}{u}} \right)} = - {\color{red}{\left(x - 5\right)}} + {\color{red}{\left(x - 5\right)}} \ln{\left({\color{red}{\left(x - 5\right)}} \right)}$$

Dus,

$$\int{\ln{\left(x - 5 \right)} d x} = - x + \left(x - 5\right) \ln{\left(x - 5 \right)} + 5$$

Vereenvoudig:

$$\int{\ln{\left(x - 5 \right)} d x} = \left(x - 5\right) \left(\ln{\left(x - 5 \right)} - 1\right)$$

Voeg de integratieconstante toe:

$$\int{\ln{\left(x - 5 \right)} d x} = \left(x - 5\right) \left(\ln{\left(x - 5 \right)} - 1\right)+C$$

$$$\int \ln\left(x - 5\right)\, dx = \left(x - 5\right) \left(\ln\left(x - 5\right) - 1\right) + C$$$A