Answer
$$2\tanh 2$$
Work Step by Step
$$\eqalign{
& \int_0^4 {\frac{{{{\operatorname{sech} }^2}\sqrt x }}{{\sqrt x }}} dx \cr
& {\text{set }}u = \sqrt x {\text{ }}\,\,\,\,\,{\text{then }}\,\,\,\,du = \frac{1}{{2\sqrt x }}dx,\,\,\,\,\,\,\,\,\,2du = \frac{1}{{\sqrt x }}dx \cr
& {\text{switch the limits of integration}} \cr
& u = \sqrt x ,\,\,\,\,\,\,\,x = 0 \to u = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 4 \to u = 2 \cr
& {\text{use the change of variable}} \cr
& \int_0^4 {\frac{{{{\operatorname{sech} }^2}\sqrt x }}{{\sqrt x }}} dx = \int_0^2 {{{\operatorname{sech} }^2}u} \left( {2du} \right) \cr
& = 2\int_0^2 {{{\operatorname{sech} }^2}u} du \cr
& {\text{integrate and evaluate the limits}} \cr
& = 2\left( {\tanh u} \right)_0^2 \cr
& = 2\left( {\tanh 2 - \tanh 0} \right) \cr
& = 2\tanh 2 \cr} $$
