Answer
$$\ln \left| {\frac{{\sqrt {{e^s} + 1} - 1}}{{\sqrt {{e^s} + 1} + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{ds}}{{\sqrt {{e^s} + 1} }}} \cr
& {\text{Integrate by using the substitution method}} \cr
& {\text{Let }}{u^2} = {e^s} + 1,\,\,\,\,\,\,{e^s} = {u^2} - 1,\,\,\,\,\,{e^s}ds = 2udu,\,\,\,\, \cr
& \,\,\,\,\,\,\,\,\,\,\,ds = \frac{{2udu}}{{{e^s}}} = \frac{{2udu}}{{{u^2} - 1}} \cr
& {\text{write the integral in terms of }}u \cr
& \int {\frac{1}{{\sqrt {{e^s} + 1} }}ds} = \int {\frac{1}{{\sqrt {{u^2}} }}} \left( {\frac{{2udu}}{{{u^2} - 1}}} \right) \cr
& = \int {\frac{2}{{{u^2} - 1}}} du = \int {\frac{2}{{\left( {u - 1} \right)\left( {u + 1} \right)}}} du \cr
& \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{2}{{\left( {u - 1} \right)\left( {u + 1} \right)}} = \frac{A}{{u - 1}} + \frac{B}{{u + 1}} \cr
& {\text{multiply by }}\left( {u - 1} \right)\left( {u + 1} \right) \cr
& 2 = A\left( {u + 1} \right) + B\left( {u - 1} \right) \cr
& \,\,\,{\text{if }}u = 1,\,\, then\,\,A = 1 \cr
& \,\,\,{\text{if }}u = - 1,\,\, then\,\,B = - 1 \cr
& \frac{1}{{\left( {u - 1} \right)\left( {u + 1} \right)}} = \frac{A}{{u - 1}} + \frac{B}{{u + 1}} = \frac{1}{{u - 1}} - \frac{1}{{u + 1}} \cr
& \cr
& {\text{Then}} \cr
& \int {\frac{2}{{\left( {u - 1} \right)\left( {u + 1} \right)}}du} = \int {\left( {\frac{1}{{u - 1}} - \frac{1}{{u + 1}}} \right)} du \cr
& = \int {\left( {\frac{1}{{u - 1}} - \frac{1}{{u + 1}}} \right)} du \cr
& {\text{integrating}} \cr
& = \left( {\ln \left| {u - 1} \right| - \ln \left| {u + 1} \right|} \right) + C \cr
& = \ln \left| {\frac{{u - 1}}{{u + 1}}} \right| + C \cr
& \cr
& {\text{Write the integral in terms of }}s,{\text{ replace }}\sqrt {{e^s} + 1} {\text{ for }}u \cr
& = \ln \left| {\frac{{\sqrt {{e^s} + 1} - 1}}{{\sqrt {{e^s} + 1} + 1}}} \right| + C \cr} $$
