Ολοκλήρωμα του $$$\tanh{\left(x \right)}$$$

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

Ξαναγράψτε την υπερβολική εφαπτομένη ως $$$\tanh\left(x\right)=\frac{\sinh\left(x\right)}{\cosh\left(x\right)}$$$:

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

Έστω $$$u=\cosh{\left(x \right)}$$$.

Τότε $$$du=\left(\cosh{\left(x \right)}\right)^{\prime }dx = \sinh{\left(x \right)} dx$$$ (τα βήματα παρουσιάζονται »), και έχουμε ότι $$$\sinh{\left(x \right)} dx = du$$$.

Επομένως,

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

Το ολοκλήρωμα του $$$\frac{1}{u}$$$ είναι $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

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

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

Επομένως,

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

Προσθέστε τη σταθερά ολοκλήρωσης:

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

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