Answer
$\frac{{11}}{6}$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{{{x^2} + x + 1}}{{{x^4}}}} dx \cr
& {\text{Distribute the numerator}} \cr
& = \int_1^\infty {\left( {\frac{{{x^2}}}{{{x^4}}} + \frac{x}{{{x^4}}} + \frac{1}{{{x^4}}}} \right)} dx \cr
& = \int_1^\infty {\left( {\frac{1}{{{x^2}}} + \frac{1}{{{x^3}}} + \frac{1}{{{x^4}}}} \right)} dx \cr
& {\text{Using the definition of improper integrals }} \cr
& \underbrace {\int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{t \to \infty } } \int_a^t {f\left( x \right)} dx}_ \Downarrow \cr
& \int_1^\infty {\left( {\frac{1}{{{x^2}}} + \frac{1}{{{x^3}}} + \frac{1}{{{x^4}}}} \right)} dx = \mathop {\lim }\limits_{t \to \infty } \int_1^t {\left( {\frac{1}{{{x^2}}} + \frac{1}{{{x^3}}} + \frac{1}{{{x^4}}}} \right)} dx \cr
& = \mathop {\lim }\limits_{t \to \infty } \int_1^t {\left( {{x^{ - 2}} + {x^{ - 3}} + {x^{ - 4}}} \right)} dx \cr
& {\text{Integrating by the power rule}} \cr
& = \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{{{x^{ - 1}}}}{{ - 1}} + \frac{{{x^{ - 2}}}}{{ - 2}} + \frac{{{x^{ - 3}}}}{{ - 3}}} \right]_1^t \cr
& = \mathop {\lim }\limits_{t \to \infty } \left[ { - \frac{1}{x} - \frac{1}{{2{x^2}}} - \frac{1}{{3{x^3}}}} \right]_1^t \cr
& = \mathop {\lim }\limits_{t \to \infty } \left( {\left[ { - \frac{1}{t} - \frac{1}{{2{t^2}}} - \frac{1}{{3{t^3}}}} \right] - \left[ { - \frac{1}{1} - \frac{1}{{2{{\left( 1 \right)}^2}}} - \frac{1}{{3{{\left( 1 \right)}^3}}}} \right]} \right) \cr
& = \mathop {\lim }\limits_{t \to \infty } \left( {\left[ { - \frac{1}{t} - \frac{1}{{2{t^2}}} - \frac{1}{{3{t^3}}}} \right] - \left[ { - \frac{{11}}{6}} \right]} \right) \cr
& = \mathop {\lim }\limits_{t \to \infty } \left( { - \frac{1}{t} - \frac{1}{{2{t^2}}} - \frac{1}{{3{t^3}}} + \frac{{11}}{6}} \right) \cr
& {\text{Evaluate the limit when }}t \to \infty \cr
& = - 0 - 0 - 0 + \frac{{11}}{6} \cr
& = \frac{{11}}{6} \cr
& {\text{Therefore}}{\text{, the integral converges.}} \cr} $$
