Answer
$\frac{1}{2}$
Work Step by Step
$$\eqalign{
& \int_0^\infty {\frac{{{e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}}} dx \cr
& {\text{Using the definition of improper integrals }} \cr
& \underbrace {\int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{t \to \infty } } \int_a^t {f\left( x \right)} dx}_ \Downarrow \cr
& \int_0^\infty {\frac{{{e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}}} dx = \mathop {\lim }\limits_{t \to \infty } \int_0^t {\frac{{{e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}}} dx \cr
& {\text{Rewrite the integrand}} \cr
& = \mathop {\lim }\limits_{t \to \infty } \int_0^t {{{\left( {1 + {e^x}} \right)}^{ - 2}}{e^x}} dx \cr
& {\text{Integrating}} \cr
& = \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{{{{\left( {1 + {e^x}} \right)}^{ - 1}}}}{{ - 1}}} \right]_0^t \cr
& = - \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{1}{{1 + {e^x}}}} \right]_0^t \cr
& = - \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{1}{{1 + {e^t}}} - \frac{1}{{1 + {e^0}}}} \right] \cr
& = - \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{1}{{1 + {e^t}}} - \frac{1}{2}} \right] \cr
& {\text{Evaluate the limit when }}t \to \infty \cr
& = - \frac{1}{{1 + {e^{\mathop {\lim }\limits_{t \to \infty } t}}}} + \frac{1}{2} \cr
& = 0 + \frac{1}{2} \cr
& = \frac{1}{2} \cr
& {\text{Therefore}}{\text{, the integral converges.}} \cr} $$
