Answer
Diverges
Work Step by Step
$$\eqalign{
& \int_{ - 2}^\infty {\frac{1}{{x + 4}}} dx \cr
& {\text{Using the definition of improper integrals }} \cr
& \underbrace {\int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{t \to \infty } } \int_a^t {f\left( x \right)} dx}_ \Downarrow \cr
& \int_{ - 2}^\infty {\frac{1}{{x + 4}}} dx = \mathop {\lim }\limits_{t \to \infty } \int_{ - 2}^t {\frac{1}{{x + 4}}} dx \cr
& {\text{Integrating}} \cr
& = \mathop {\lim }\limits_{t \to \infty } \left[ {\ln \left| {x + 4} \right|} \right]_{ - 2}^t \cr
& = \mathop {\lim }\limits_{t \to \infty } \left[ {\ln \left| {t + 4} \right| - \ln \left| { - 2 + 4} \right|} \right] \cr
& = \mathop {\lim }\limits_{t \to \infty } \left[ {\ln \left| {t + 4} \right| - \ln \left| 2 \right|} \right] \cr
& = \mathop {\lim }\limits_{t \to \infty } \left[ {\ln \left| {t + 4} \right|} \right] - \mathop {\lim }\limits_{t \to \infty } \left[ {\ln \left| 2 \right|} \right] \cr
& {\text{Evaluate the limit when }}t \to \infty \cr
& = \infty - \ln 2 \cr
& = \infty \cr
& {\text{Therefore}}{\text{, the integral diverges}} \cr} $$
