Integraal van $$$\ln\left(f x\right)$$$ met betrekking tot $$$x$$$

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

Zij $$$u=f x$$$.

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

Dus,

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

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

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

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

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

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

De integraal wordt

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

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

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

We herinneren eraan dat $$$u=f x$$$:

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

Dus,

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

Vereenvoudig:

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

Voeg de integratieconstante toe:

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

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