# Antiderivative of tanh

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