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

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

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

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

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

Das Integral wird zu

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

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=-2$$$ an:

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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