Answer
$\frac{\pi }{4} - \frac{1}{2}\arctan \left( {\frac{1}{2}} \right)$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{1}{{{x^2} + 4}}} dx \cr
& {\text{Using the definition of improper integrals }} \cr
& \underbrace {\int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{t \to \infty } } \int_a^t {f\left( x \right)} dx}_ \Downarrow \cr
& \int_1^\infty {\frac{1}{{{x^2} + 4}}} dx = \mathop {\lim }\limits_{t \to \infty } \int_1^t {\frac{1}{{{x^2} + 4}}} dx \cr
& = \mathop {\lim }\limits_{t \to \infty } \int_1^t {\frac{1}{{{x^2} + {{\left( 2 \right)}^2}}}} dx \cr
& {\text{Integrating}}{\text{, apply}}\int {\frac{1}{{{x^2} + {a^2}}}dx = \frac{1}{a}\arctan \left( {\frac{x}{a}} \right) + C} \cr
& = \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{1}{2}\arctan \left( {\frac{x}{2}} \right)} \right]_1^t \cr
& = \frac{1}{2}\mathop {\lim }\limits_{t \to \infty } \left[ {\arctan \left( {\frac{x}{2}} \right)} \right]_1^t \cr
& = \frac{1}{2}\mathop {\lim }\limits_{t \to \infty } \left[ {\arctan \left( {\frac{t}{2}} \right) - \arctan \left( {\frac{1}{2}} \right)} \right] \cr
& {\text{Evaluate the limit when }}t \to \infty \cr
& = \frac{1}{2}\left( {\frac{\pi }{2}} \right) - \frac{1}{2}\arctan \left( {\frac{1}{2}} \right) \cr
& = \frac{\pi }{4} - \frac{1}{2}\arctan \left( {\frac{1}{2}} \right) \cr
& {\text{Therefore}}{\text{, the integral converges.}} \cr} $$
