Answer
$$\sqrt {1 + {e^2}} - \sqrt 2 $$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{{e^{2z}}}}{{\sqrt {1 + {e^{2z}}} }}} dz \cr
& {\text{use substitution}}{\text{. Let }}u = 1 + {e^{2z}},{\text{ so that }}du = 2{e^{2z}},\,\,\,\,\,\,\,\frac{1}{2}du = {e^{2z}}du \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 1,{\text{ then }}u = 1 + {e^{2\left( 1 \right)}} = 1 + {e^2} \cr
& \,\,\,\,\,\,{\text{If }}x = 0,{\text{ then }}u = 1 + {e^{2\left( 0 \right)}} = 2 \cr
& {\text{Then}} \cr
& \int_0^1 {\frac{{{e^{2z}}}}{{\sqrt {1 + {e^{2z}}} }}} dz = \int_2^{1 + {e^2}} {\frac{1}{{\sqrt u }}\left( {\frac{1}{2}du} \right)} \cr
& = \frac{1}{2}\int_2^{1 + {e^2}} {\frac{1}{{\sqrt u }}du} \cr
& = \frac{1}{2}\int_2^{1 + {e^2}} {\frac{1}{{{u^{1/2}}}}du} \cr
& = \frac{1}{2}\int_2^{1 + {e^2}} {{u^{ - 1/2}}du} \cr
& {\text{use }}\int_a^b {{u^n}} du = \left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right)_a^b \cr
& = \frac{1}{2}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right)_2^{1 + {e^2}} \cr
& = \left( {{u^{1/2}}} \right)_2^{1 + {e^2}} \cr
& = \left( {\sqrt u } \right)_2^{1 + {e^2}} \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \sqrt {1 + {e^2}} - \sqrt 2 \cr} $$
