Answer
$${\text{The integral is divergent}}$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{{4x + 6}}{{{x^2} + 3x}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{}} \cr
& \int_1^\infty {\frac{{4x + 6}}{{{x^2} + 3x}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{{4x + 6}}{{{x^2} + 3x}}} dx \cr
& \cr
& {\text{integrating }}\int {\frac{{4x + 6}}{{{x^2} + 3x}}} dx \cr
& {\text{set }}u = {x^2} + 3x{\text{ then }}\frac{{du}}{{dx}} = 2x + 3,\,\,\,\,\frac{{du}}{{2x + 3}} = dx \cr
& \int {\frac{{4x + 6}}{{{x^2} + 3x}}} dx = \int {\frac{{2\left( {2x + 3} \right)}}{u}} dx\left( {\frac{{du}}{{2x + 3}}} \right) = 2\int {\frac{{du}}{u}} \cr
& = 2\ln \left| u \right| + C \cr
& {\text{replace }}u = {x^2} + 3x \cr
& = 2\ln \left| {{x^2} + 3x} \right| + C \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{{4x + 6}}{{{x^2} + 3x}}} dx = 2\mathop {\lim }\limits_{b \to \infty } \left( {\ln \left| {{x^2} + 3x} \right|} \right)_1^b \cr
& = 2\mathop {\lim }\limits_{a \to - \infty } \left( {\ln \left| {{b^2} + 3b} \right| - \ln \left| {{1^2} + 3\left( 1 \right)} \right|} \right) \cr
& = 2\mathop {\lim }\limits_{a \to - \infty } \left( {\ln \left| {{b^2} + 3b} \right| - \ln 4} \right) \cr
& {\text{evaluating the limit when }}b \to \infty \cr
& = 2\left( {\ln \left| \infty \right| - \ln 4} \right) \cr
& = \infty \cr
& then \cr
& {\text{The integral is divergent}} \cr} $$
