Answer
$$\frac{1}{{2\left( {{a^2} + 1} \right)}}$$
Work Step by Step
$$\eqalign{
& \int_a^{ + \infty } {\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx \cr
& {\text{using the definition 7}}{\text{.8}}{\text{.1 of improper integrals}} \cr
& \,\,\,\int_a^{ + \infty } {f\left( x \right)} dx = \mathop {\lim }\limits_{b \to + \infty } \int_a^b {f\left( x \right)} dx \cr
& \cr
& {\text{then}} \cr
& \,\,\int_a^{ + \infty } {\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx = \mathop {\lim }\limits_{b \to + \infty } \int_a^b {\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx \cr
& = \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \int_a^b {\frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx \cr
& \cr
& {\text{Integrate}} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ { - \frac{1}{{{x^2} + 1}}} \right]_a^b \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{{{b^2} + 1}} - \frac{1}{{{a^2} + 1}}} \right] \cr
& \cr
& {\text{calculate the limit when }}b \to + \infty \cr
& = - \frac{1}{2}\left[ {\frac{1}{{{{\left( \infty \right)}^2} + 1}} - \frac{1}{{{a^2} + 1}}} \right] \cr
& {\text{simplifying}} \cr
& = - \frac{1}{2}\left[ {\frac{1}{\infty } - \frac{1}{{{a^2} + 1}}} \right] \cr
& = - \frac{1}{2}\left( { - \frac{1}{{{a^2} + 1}}} \right) \cr
& = \frac{1}{{2\left( {{a^2} + 1} \right)}} \cr
& \cr
& {\text{then}}{\text{,}} \cr
& {\text{The integral converges to }}\frac{1}{{2\left( {{a^2} + 1} \right)}} \cr} $$
