Answer
$${\text{diverges}}$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{{3{x^2} + 1}}{{{x^3} + x}}} dx \cr
& {\text{definition of improper integral}} \cr
& \int_1^\infty {\frac{{3{x^2} + 1}}{{{x^3} + x}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{{3{x^2} + 1}}{{{x^3} + x}}} dx \cr
& {\text{evaluate the integral}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left. {\left( {\ln \left( {{x^3} + x} \right)} \right)} \right|_1^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( {\ln \left( {{b^3} + b} \right) - \ln \left( {{1^3} + 1} \right)} \right) \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( {\ln \left( {{b^3} + b} \right) - \ln \left( 2 \right)} \right) \cr
& {\text{evaluate the limit}} \cr
& = \left( {\ln \left( {{\infty ^3} + \infty } \right) - \ln \left( 2 \right)} \right) \cr
& = \infty - \ln 2 \cr
& = b \cr
& {\text{diverges}} \cr} $$
