Answer
$$\frac{1}{{32}}$$
Work Step by Step
$$\eqalign{
& \int_3^\infty {\frac{1}{{{{\left( {x + 1} \right)}^3}}}} dx \cr
& {\text{write with negative exponent}} \cr
& = \int_3^\infty {{{\left( {x + 1} \right)}^{ - 3}}} dx \cr
& {\text{solve the improper integral using the definition }}\int_a^\infty {f\left( x \right)} dx = \mathop {\lim }\limits_{b \to \infty } \int_a^b {f\left( x \right)} dx{\text{ }} \cr
& {\text{then}} \cr
& \int_3^\infty {{{\left( {x + 1} \right)}^{ - 3}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_3^b {{{\left( {x + 1} \right)}^{ - 3}}} dx{\text{ }} \cr
& {\text{integrate by using the power property }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& \int_3^\infty {{{\left( {x + 1} \right)}^{ - 3}}} dx = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{{\left( {x + 1} \right)}^{ - 2}}}}{{ - 2}}} \right]_3^b \cr
& \int_3^\infty {{{\left( {x + 1} \right)}^{ - 3}}} dx = - \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right]_3^b \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& \int_3^\infty {{{\left( {x + 1} \right)}^{ - 3}}} dx = - \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{{{\left( {b + 1} \right)}^2}}} - \frac{1}{{{{\left( {3 + 1} \right)}^2}}}} \right] \cr
& \int_3^\infty {{{\left( {x + 1} \right)}^{ - 3}}} dx = - \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{{{\left( {b + 1} \right)}^2}}} - \frac{1}{{16}}} \right] \cr
& {\text{evaluate the limit when }}b \to \infty \cr
& \int_3^\infty {{{\left( {x + 1} \right)}^{ - 3}}} dx = - \frac{1}{2}\left[ {\frac{1}{{{{\left( {\infty + 1} \right)}^2}}} - \frac{1}{{16}}} \right] \cr
& {\text{Simplify}} \cr
& \int_3^\infty {{{\left( {x + 1} \right)}^{ - 3}}} dx = - \frac{1}{2}\left( { - \frac{1}{{16}}} \right) \cr
& \int_3^\infty {{{\left( {x + 1} \right)}^{ - 3}}} dx = \frac{1}{{32}} \cr} $$
