# Antiderivative of coth

The following formula holds: $$\displaystyle\int \mathrm{coth}(z) \mathrm{d}z=\log(\sinh(z)) + C,$$ for arbitrary constant $C$, where $\mathrm{coth}$ denotes the hyperbolic cotangent, $\log$ denotes the logarithm, and $\sinh$ denotes the hyperbolic sine.
By definition, $$\mathrm{coth}(z) = \dfrac{\mathrm{cosh}(z)}{\mathrm{sinh}(z)}.$$ Let $u=\mathrm{sinh}(z)$ and use the derivative of sinh, u-substitution, and the definition of the logarithm to derive $$\begin{array}{ll} \displaystyle\int \mathrm{coth}(z) \mathrm{d}z &= \displaystyle\int \dfrac{1}{u} \mathrm{d} u \\ &= \log \left( \mathrm{sinh}(z) \right) + C, \end{array}$$ as was to be shown. █