Answer
$$\frac{{{{\left( {\tanh x} \right)}^2}}}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\operatorname{sech} }^2}x\tanh x} dx \cr
& {\text{substitute }}u = \tanh x,{\text{ }}du = {\operatorname{sech} ^2}xdx \cr
& \int {{{\operatorname{sech} }^2}x\tanh x} dx = \int u du \cr
& {\text{find the antiderivative by the power rule}} \cr
& = \frac{{{u^2}}}{2} + C \cr
& {\text{replacing}}\,\,\,u = \tanh x \cr
& = \frac{{{{\left( {\tanh x} \right)}^2}}}{2} + C \cr} $$
