Integral von $$$\frac{1}{x \ln\left(x\right)}$$$

Bestimme $$$\int \frac{1}{x \ln\left(x\right)}\, dx$$$.

Sei $$$u=\ln{\left(x \right)}$$$.

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

Daher,

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

Das Integral von $$$\frac{1}{u}$$$ ist $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$${\color{red}{\int{\frac{1}{u} d u}}} = {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Zur Erinnerung: $$$u=\ln{\left(x \right)}$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\ln{\left(x \right)}}}}\right| \right)}$$

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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