Answer
$$2\left( {x - 2} \right)\sqrt {1 + {e^x}} + 2\ln \left| {\frac{{\sqrt {1 + {e^x}} - 1}}{{\sqrt {1 + {e^x}} + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{x{e^x}}}{{\sqrt {1 + {e^x}} }}} dx \cr
& {\text{Rewrite}} \cr
& = \int {x\left( {\frac{{{e^x}}}{{\sqrt {1 + {e^x}} }}} \right)} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = x,{\text{ }}du = dx \cr
& dv = \frac{{{e^x}}}{{\sqrt {1 + {e^x}} }}dx \to v = 2\sqrt {1 + {e^x}} \cr
& {\text{Use integration by parts formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {\frac{{x{e^x}}}{{\sqrt {1 + {e^x}} }}} dx = 2x\sqrt {1 + {e^x}} - 2\int {\sqrt {1 + {e^x}} } dx{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{*Integrating }}\int {\sqrt {1 + {e^x}} } dx \cr
& {\text{Let }}{t^2} = 1 + {e^x} \to 2tdt = {e^x}dx \cr
& {\text{ }}{t^2} = 1 + {e^x} \to {e^x} = {t^2} - 1 \to x = \ln \left( {{t^2} - 1} \right) \cr
& \int {\sqrt {1 + {e^x}} } dx = \int {t\left( {\frac{{2t}}{{{e^x}}}} \right)} dt \cr
& = \int {\frac{{2{t^2}}}{{{t^2} - 1}}} dt = \int {\left( {2 - \frac{2}{{{t^2} - 1}}} \right)} dt \cr
& Integrating \cr
& = 2t - \ln \left| {\frac{{t + 1}}{{t - 1}}} \right| + C \cr
& {\text{Write in terms of }}x \cr
& = 2\sqrt {1 + {e^x}} - \ln \left| {\frac{{\sqrt {1 + {e^x}} + 1}}{{\sqrt {1 + {e^x}} - 1}}} \right| + C \cr
& {\text{Substituting the previous result into }}\left( {\bf{1}} \right) \cr
& = 2x\sqrt {1 + {e^x}} - 2\left( {2\sqrt {1 + {e^x}} - \ln \left| {\frac{{\sqrt {1 + {e^x}} + 1}}{{\sqrt {1 + {e^x}} - 1}}} \right|} \right) + C \cr
& = 2x\sqrt {1 + {e^x}} - 4\sqrt {1 + {e^x}} + 2\ln \left| {\frac{{\sqrt {1 + {e^x}} + 1}}{{\sqrt {1 + {e^x}} + 1}}} \right| + C \cr
& = 2x\sqrt {1 + {e^x}} - 4\sqrt {1 + {e^x}} + 2\ln \left| {\frac{{\sqrt {1 + {e^x}} - 1}}{{\sqrt {1 + {e^x}} + 1}}} \right| + C \cr
& {\text{Factoring}} \cr
& = 2\left( {x - 2} \right)\sqrt {1 + {e^x}} + 2\ln \left| {\frac{{\sqrt {1 + {e^x}} - 1}}{{\sqrt {1 + {e^x}} + 1}}} \right| + C \cr} $$
