Integral of $$$\coth{\left(x \right)}$$$

Rewrite the hyperbolic cotangent as $$$\coth\left(x\right)=\frac{\cosh\left(x\right)}{\sinh\left(x\right)}$$$:

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

Let $$$u=\sinh{\left(x \right)}$$$.

Then $$$du=\left(\sinh{\left(x \right)}\right)^{\prime }dx = \cosh{\left(x \right)} dx$$$ (steps can be seen »), and we have that $$$\cosh{\left(x \right)} dx = du$$$.

Thus,

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

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Recall that $$$u=\sinh{\left(x \right)}$$$:

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

Therefore,

$$\int{\coth{\left(x \right)} d x} = \ln{\left(\left|{\sinh{\left(x \right)}}\right| \right)}$$

Add the constant of integration:

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

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