Answer
$$\infty \,\,\,\left( {{\text{diverges}}} \right)$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^{ + \infty } {\frac{x}{{1 + {x^2}}}} dx \cr
& {\text{using the definition 7}}{\text{.8}}{\text{.1 of improper integrals}} \cr
& \,\,\,\int_a^{ + \infty } {f\left( x \right)} dx = \mathop {\lim }\limits_{b \to + \infty } \int_a^b {f\left( x \right)} dx \cr
& \cr
& {\text{then}} \cr
& \,\,\,\int_{ - 1}^{ + \infty } {\frac{x}{{1 + {x^2}}}} dx = \mathop {\lim }\limits_{b \to + \infty } \int_{ - 1}^b {\frac{x}{{1 + {x^2}}}} dx \cr
& {\text{integrating}} \cr
& = \mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{2}\ln \left( {1 + {x^2}} \right)} \right]_{ - 1}^b \cr
& = \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {\ln \left( {1 + {b^2}} \right) + \ln \left( {1 + {{\left( { - 1} \right)}^2}} \right)} \right] \cr
& = \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {\ln \left( {1 + {b^2}} \right) + \ln \left( 2 \right)} \right] \cr
& \cr
& {\text{calculate the limit when }}b \to + \infty \cr
& = \frac{1}{2}\left[ {\ln \left( {1 + {{\left( { + \infty } \right)}^2}} \right) + \ln \left( 2 \right)} \right] \cr
& = \frac{1}{2}\left[ {\ln \left( \infty \right) + \ln \left( 2 \right)} \right] \cr
& = \infty \cr
& {\text{In this case the integral diverges}} \cr} $$
