Answer
$$\infty $$
Work Step by Step
$$\eqalign{
& \int_0^\infty {\frac{{{e^x}}}{{1 + {e^x}}}dx} \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ so}} \cr
& \int_0^\infty {\frac{{{e^x}}}{{1 + {e^x}}}dx} = \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{{{e^x}}}{{1 + {e^x}}}dx} \cr
& {\text{Integrating}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\ln \left( {1 + {e^x}} \right)} \right]_0^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\ln \left( {1 + {e^b}} \right) - \ln \left( {1 + {e^0}} \right)} \right] \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\ln \left( {1 + {e^b}} \right) - \ln \left( 2 \right)} \right] \cr
& {\text{Evaluate the limit}} \cr
& = \ln \left( {1 + {e^\infty }} \right) - \ln \left( 2 \right) \cr
& = \infty \cr} $$
