Answer
$$ - \ln \left( {\frac{{1 + \sqrt {1 + {e^{2x}}} }}{{{e^x}}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{\sqrt {1 + {e^{2x}}} }}} dx \cr
& {\text{Let }}u = {e^x},{\text{ }}du = {e^x}dx,{\text{ }}dx = \frac{1}{{{e^x}}}du = \frac{1}{u}du \cr
& {\text{Substitute}} \cr
& \int {\frac{1}{{\sqrt {1 + {e^{2x}}} }}} dx = \int {\frac{1}{{\sqrt {1 + {u^2}} }}} \left( {\frac{1}{u}} \right)du \cr
& = \int {\frac{1}{{u\sqrt {1 + {u^2}} }}} du \cr
& {\text{Use Theorem 5}}{\text{.20 }}\int {\frac{{du}}{{u\sqrt {{a^2} + {u^2}} }} = - \frac{1}{a}\ln \frac{{a + \sqrt {{a^2} + {u^2}} }}{{\left| u \right|}} + C} \cr
& \int {\frac{1}{{u\sqrt {1 + {u^2}} }}} du = - \ln \left( {\frac{{1 + \sqrt {1 + {u^2}} }}{{\left| u \right|}}} \right) + C \cr
& {\text{Write in terms of }}x \cr
& = - \ln \left( {\frac{{1 + \sqrt {1 + {{\left( {{e^x}} \right)}^2}} }}{{\left| {{e^x}} \right|}}} \right) + C \cr
& = - \ln \left( {\frac{{1 + \sqrt {1 + {e^{2x}}} }}{{{e^x}}}} \right) + C \cr} $$
