Integral von $$$\tanh{\left(x \right)}$$$

Bestimme $$$\int \tanh{\left(x \right)}\, dx$$$.

Schreibe den hyperbolischen Tangens als $$$\tanh\left(x\right)=\frac{\sinh\left(x\right)}{\cosh\left(x\right)}$$$ um.:

$${\color{red}{\int{\tanh{\left(x \right)} d x}}} = {\color{red}{\int{\frac{\sinh{\left(x \right)}}{\cosh{\left(x \right)}} d x}}}$$

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

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

Somit,

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

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: $$$u=\cosh{\left(x \right)}$$$:

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

Daher,

$$\int{\tanh{\left(x \right)} d x} = \ln{\left(\cosh{\left(x \right)} \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\tanh{\left(x \right)} d x} = \ln{\left(\cosh{\left(x \right)} \right)}+C$$

$$$\int \tanh{\left(x \right)}\, dx = \ln\left(\cosh{\left(x \right)}\right) + C$$$A