Answer
$$\ln \left| {\frac{{\sqrt {{e^x} + 1} - 1}}{{\sqrt {{e^x} + 1} + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{\sqrt {{e^x} + 1} }}} dx \cr
& \cr
& {\text{Integrate by the substitution method}} \cr
& {u^2} = {e^x} + 1,\,\,\,\,\,\,\,\,\,\,\,2udu = {e^x}dx,\,\,\,\,\,\,\,\,\,\,dx = \frac{{2udu}}{{{e^x}}} = \frac{{2udu}}{{{u^2} - 1}} \cr
& {\text{Write the integrand in terms of }}u \cr
& \int {\frac{1}{{\sqrt {{e^x} + 1} }}} dx = \int {\frac{1}{{\sqrt {{u^2}} }}} \left( {\frac{{2udu}}{{{u^2} - 1}}} \right) \cr
& = \int {\frac{1}{u}} \left( {\frac{{2udu}}{{{u^2} - 1}}} \right) \cr
& = 2\int {\frac{1}{{{u^2} - 1}}} du \cr
& {\text{Use the endpaper integration formulas }}\int {\frac{1}{{{u^2} - {a^2}}}du = \frac{1}{{2a}}\ln \left| {\frac{{u - a}}{{u + a}}} \right| + C} \cr
& = 2\left( {\frac{1}{{2\left( 1 \right)}}\ln \left| {\frac{{u - 1}}{{u + 1}}} \right|} \right) + C \cr
& = \ln \left| {\frac{{u - 1}}{{u + 1}}} \right| + C \cr
& \cr
& {\text{Write the integrand in terms of }}x,{\text{ replace }}\sqrt {{e^x} + 1} {\text{ for }}u \cr
& = \ln \left| {\frac{{\sqrt {{e^x} + 1} - 1}}{{\sqrt {{e^x} + 1} + 1}}} \right| + C \cr} $$
