Answer
$${\text{ln}}\left| {\tanh x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\operatorname{sech} }^2}x}}{{\tanh x}}dx} \cr
& {\text{Let }}u = \tanh x,{\text{ }}du = {\operatorname{sech} ^2}xdx \cr
& {\text{Substitute}} \cr
& \int {\frac{{{{\operatorname{sech} }^2}x}}{{\tanh x}}dx} = \int {\frac{{du}}{u}} \cr
& {\text{Integrating}} \cr
& = {\text{ln}}\left| u \right| + C \cr
& {\text{Write in terms of }}x \cr
& = {\text{ln}}\left| {\tanh x} \right| + C \cr} $$
