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

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

Sei $$$u=\frac{1}{x}$$$.

Dann $$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (die Schritte sind » zu sehen), und es gilt $$$\frac{dx}{x^{2}} = - du$$$.

Daher,

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

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

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

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

Seien $$$\operatorname{m}=\ln{\left(u \right)}^{2}$$$ und $$$\operatorname{dv}=du$$$.

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

Also,

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

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=2$$$ und $$$f{\left(u \right)} = \ln{\left(u \right)}$$$ an:

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

Für das Integral $$$\int{\ln{\left(u \right)} d u}$$$ verwenden Sie die partielle Integration $$$\int \operatorname{m} \operatorname{dv} = \operatorname{m}\operatorname{v} - \int \operatorname{v} \operatorname{dm}$$$.

Seien $$$\operatorname{m}=\ln{\left(u \right)}$$$ und $$$\operatorname{dv}=du$$$.

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

Das Integral wird zu

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

Wenden Sie die Konstantenregel $$$\int c\, du = c u$$$ mit $$$c=1$$$ an:

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

Zur Erinnerung: $$$u=\frac{1}{x}$$$:

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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