Integral von $$$\frac{\ln\left(x^{2}\right)}{x^{2}}$$$

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

Die Eingabe wird umgeschrieben: $$$\int{\frac{\ln{\left(x^{2} \right)}}{x^{2}} d x}=\int{\frac{2 \ln{\left(x \right)}}{x^{2}} d x}$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=2$$$ und $$$f{\left(x \right)} = \frac{\ln{\left(x \right)}}{x^{2}}$$$ an:

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

Für das Integral $$$\int{\frac{\ln{\left(x \right)}}{x^{2}} d x}$$$ verwenden Sie die partielle Integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

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

Dann gilt $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (Rechenschritte siehe ») und $$$\operatorname{v}=\int{\frac{1}{x^{2}} d x}=- \frac{1}{x}$$$ (Rechenschritte siehe »).

Also,

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

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=-1$$$ und $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$ an:

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

Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=-2$$$ an:

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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