Answer
$$ = \frac{2}{3}{\left( {\tanh x} \right)^{3/2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {\tanh x} {{\operatorname{sech} }^2}x} dx \cr
& {\text{substitute }}u = \tanh x,{\text{ }}du = {\operatorname{sech} ^2}xdx \cr
& = \int {\sqrt {\tanh x} {{\operatorname{sech} }^2}x} dx = \int {\sqrt u du} \cr
& = \int {{u^{1/2}}du} \cr
& {\text{find the antiderivarive using the power rule}} \cr
& = \frac{{{u^{3/2}}}}{{3/2}} + C \cr
& = \frac{2}{3}{u^{3/2}} + C \cr
& {\text{write in terms of }}x \cr
& = \frac{2}{3}{\left( {\tanh x} \right)^{3/2}} + C \cr} $$
