Integraal van $$$\frac{\ln\left(x^{2}\right)}{x^{2}}$$$

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

De invoer is herschreven: $$$\int{\frac{\ln{\left(x^{2} \right)}}{x^{2}} d x}=\int{\frac{2 \ln{\left(x \right)}}{x^{2}} d x}$$$.

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

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

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

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

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

Dus,

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=-1$$$ en $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$:

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

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=-2$$$:

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

Dus,

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

Vereenvoudig:

$$\int{\frac{2 \ln{\left(x \right)}}{x^{2}} d x} = \frac{2 \left(- \ln{\left(x \right)} - 1\right)}{x}$$

Voeg de integratieconstante toe:

$$\int{\frac{2 \ln{\left(x \right)}}{x^{2}} d x} = \frac{2 \left(- \ln{\left(x \right)} - 1\right)}{x}+C$$

$$$\int \frac{\ln\left(x^{2}\right)}{x^{2}}\, dx = \frac{2 \left(- \ln\left(x\right) - 1\right)}{x} + C$$$A