Integraal van $$$- 4 \ln^{2}\left(x\right) + 2 \ln\left(x\right)$$$

Bepaal $$$\int \left(- 4 \ln^{2}\left(x\right) + 2 \ln\left(x\right)\right)\, dx$$$.

Integreer termgewijs:

$${\color{red}{\int{\left(- 4 \ln{\left(x \right)}^{2} + 2 \ln{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{2 \ln{\left(x \right)} d x} - \int{4 \ln{\left(x \right)}^{2} d x}\right)}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=4$$$ en $$$f{\left(x \right)} = \ln{\left(x \right)}^{2}$$$:

$$\int{2 \ln{\left(x \right)} d x} - {\color{red}{\int{4 \ln{\left(x \right)}^{2} d x}}} = \int{2 \ln{\left(x \right)} d x} - {\color{red}{\left(4 \int{\ln{\left(x \right)}^{2} d x}\right)}}$$

Voor de integraal $$$\int{\ln{\left(x \right)}^{2} d x}$$$, gebruik partiële integratie $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Zij $$$\operatorname{u}=\ln{\left(x \right)}^{2}$$$ en $$$\operatorname{dv}=dx$$$.

Dan $$$\operatorname{du}=\left(\ln{\left(x \right)}^{2}\right)^{\prime }dx=\frac{2 \ln{\left(x \right)}}{x} dx$$$ (de stappen zijn te zien ») en $$$\operatorname{v}=\int{1 d x}=x$$$ (de stappen zijn te zien »).

Dus,

$$\int{2 \ln{\left(x \right)} d x} - 4 {\color{red}{\int{\ln{\left(x \right)}^{2} d x}}}=\int{2 \ln{\left(x \right)} d x} - 4 {\color{red}{\left(\ln{\left(x \right)}^{2} \cdot x-\int{x \cdot \frac{2 \ln{\left(x \right)}}{x} d x}\right)}}=\int{2 \ln{\left(x \right)} d x} - 4 {\color{red}{\left(x \ln{\left(x \right)}^{2} - \int{2 \ln{\left(x \right)} d x}\right)}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=2$$$ en $$$f{\left(x \right)} = \ln{\left(x \right)}$$$:

$$- 4 x \ln{\left(x \right)}^{2} + 5 {\color{red}{\int{2 \ln{\left(x \right)} d x}}} = - 4 x \ln{\left(x \right)}^{2} + 5 {\color{red}{\left(2 \int{\ln{\left(x \right)} d x}\right)}}$$

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

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

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

De integraal wordt

$$- 4 x \ln{\left(x \right)}^{2} + 10 {\color{red}{\int{\ln{\left(x \right)} d x}}}=- 4 x \ln{\left(x \right)}^{2} + 10 {\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}=- 4 x \ln{\left(x \right)}^{2} + 10 {\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}$$

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=1$$$:

$$- 4 x \ln{\left(x \right)}^{2} + 10 x \ln{\left(x \right)} - 10 {\color{red}{\int{1 d x}}} = - 4 x \ln{\left(x \right)}^{2} + 10 x \ln{\left(x \right)} - 10 {\color{red}{x}}$$

Dus,

$$\int{\left(- 4 \ln{\left(x \right)}^{2} + 2 \ln{\left(x \right)}\right)d x} = - 4 x \ln{\left(x \right)}^{2} + 10 x \ln{\left(x \right)} - 10 x$$

Vereenvoudig:

$$\int{\left(- 4 \ln{\left(x \right)}^{2} + 2 \ln{\left(x \right)}\right)d x} = 2 x \left(- 2 \ln{\left(x \right)}^{2} + 5 \ln{\left(x \right)} - 5\right)$$

Voeg de integratieconstante toe:

$$\int{\left(- 4 \ln{\left(x \right)}^{2} + 2 \ln{\left(x \right)}\right)d x} = 2 x \left(- 2 \ln{\left(x \right)}^{2} + 5 \ln{\left(x \right)} - 5\right)+C$$

$$$\int \left(- 4 \ln^{2}\left(x\right) + 2 \ln\left(x\right)\right)\, dx = 2 x \left(- 2 \ln^{2}\left(x\right) + 5 \ln\left(x\right) - 5\right) + C$$$A