Answer
$$\ln \left| {\frac{{\sqrt {1 + {e^x} + {e^{2x}}} }}{2} + \frac{{{e^x} + 1/2}}{{\sqrt 3 /2}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x}}}{{\sqrt {1 + {e^x} + {e^{2x}}} }}} dx \cr
& {\text{completing the square for }}1 + {e^x} + {e^{2x}} \cr
& 1 + {e^x} + {e^{2x}} = {e^{2x}} + {e^x} + \frac{1}{4} + \frac{3}{4} \cr
& 1 + {e^x} + {e^{2x}} = {\left( {{e^x} + \frac{1}{2}} \right)^2} + \frac{3}{4} \cr
& \int {\frac{{{e^x}}}{{\sqrt {1 + {e^x} + {e^{2x}}} }}} dx = \int {\frac{{{e^x}}}{{\sqrt {{{\left( {{e^x} + 1/2} \right)}^2} + 3/4} }}} dx \cr
& \cr
& {\text{Write in terms of }}\theta \cr
& {\text{substitute }}{e^x} + \frac{1}{2} = \frac{{\sqrt 3 }}{2}\tan \theta ,{\text{ }}\,\,{e^x}\,dx = \frac{{\sqrt 3 }}{2}{\sec ^2}\theta d\theta \cr
& \int {\frac{{{e^x}}}{{\sqrt {{{\left( {{e^x} + 1/2} \right)}^2} + 3/4} }}} dx = \int {\frac{{\left( {\sqrt 3 /2} \right){{\sec }^2}\theta d\theta }}{{\sqrt {{{\left( {\left( {\sqrt 3 /2} \right)\tan \theta - 1/2 + 1/2} \right)}^2} + 3/4} }}} \cr
& = \int {\frac{{\left( {\sqrt 3 /2} \right){{\sec }^2}\theta d\theta }}{{\sqrt {{{\left( {\left( {\sqrt 3 /2} \right)\tan \theta } \right)}^2} + 3/4} }}} \cr
& = \int {\frac{{\left( {\sqrt 3 /2} \right){{\sec }^2}\theta d\theta }}{{\sqrt {\left( {3/4} \right){{\tan }^2}\theta + 3/4} }}} \cr
& = \int {\frac{{{{\sec }^2}\theta d\theta }}{{\sqrt {{{\tan }^2}\theta + 1} }}} \cr
& = \int {\frac{{{{\sec }^2}\theta d\theta }}{{\sqrt {{{\sec }^2}\theta } }}} \cr
& = \int {\frac{{{{\sec }^2}\theta d\theta }}{{\sec \theta }}} \cr
& = \int {\sec \theta } d\theta \cr
& {\text{Integrate}} \cr
& = \ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& \cr
& {\text{Where }}\sec \theta = \frac{{\sqrt {1 + {e^x} + {e^{2x}}} }}{2}{\text{,}}\,{\text{ and tan}}\theta = \frac{{{e^x} + 1/2}}{{\sqrt 3 /2}} \cr
& = \ln \left| {\frac{{\sqrt {1 + {e^x} + {e^{2x}}} }}{2} + \frac{{{e^x} + 1/2}}{{\sqrt 3 /2}}} \right| + C \cr} $$
