Answer
$$2\sqrt {{e^x} + 1} + \ln \left| {\sqrt {{e^x} + 1} - 1} \right| - \ln \left| {\sqrt {{e^x} + 1} + 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {{e^x} + 1} dx} \cr
& {\text{set }}u = \sqrt {{e^x} + 1} ,{\text{ }}du = \frac{{{e^x}}}{{2\sqrt {1 + {e^x}} }}dx,{\text{ }}dx = \frac{{2\sqrt {1 + {e^x}} }}{{{e^x}}} \cr
& dx = \frac{{2u}}{{{e^x}}}du \cr
& u = \sqrt {{e^x} + 1} ,{\text{ }}{e^x} = {u^2} - 1 \cr
& dx = \frac{{2u}}{{{u^2} - 1}}du \cr
& {\text{substitute}} \cr
& \int {\sqrt {{e^x} + 1} dx} = \int u \left( {\frac{{2u}}{{{u^2} - 1}}du} \right) \cr
& = \int {\frac{{2{u^2}}}{{{u^2} - 1}}du} \cr
& = \int {\left( {2 + \frac{2}{{{u^2} - 1}}} \right)du} \cr
& {\text{partial fractions}} \cr
& \frac{2}{{{u^2} - 1}} = \frac{2}{{\left( {u + 1} \right)\left( {u - 1} \right)}} \cr
& \frac{2}{{\left( {u + 1} \right)\left( {u - 1} \right)}} = \frac{A}{{u + 1}} + \frac{B}{{u - 1}} \cr
& 2 = A\left( {u - 1} \right) + B\left( {u + 1} \right) \cr
& u = - 1 \to - 1 = A \cr
& u = 1 \to 1 = B \cr
& \frac{2}{{\left( {u + 1} \right)\left( {u - 1} \right)}} = - \frac{1}{{u + 1}} + \frac{1}{{u - 1}} \cr
& = \int {\left( {2 + \frac{2}{{{u^2} - 1}}} \right)du} = \int {\left( {2 - \frac{1}{{u + 1}} + \frac{1}{{u - 1}}} \right)du} \cr
& {\text{Integrate}} \cr
& = 2u - \ln \left| {u + 1} \right| + \ln \left| {u - 1} \right| + C \cr
& with\,\,u = \sqrt {{e^x} + 1} \cr
& = 2\sqrt {{e^x} + 1} + \ln \left| {\sqrt {{e^x} + 1} - 1} \right| - \ln \left| {\sqrt {{e^x} + 1} + 1} \right| + C \cr} $$
