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

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

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

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

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

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

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

Das Integral wird zu

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

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=1$$$ an:

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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