Answer
$$\frac{1}{8}$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {\frac{{dx}}{{{{\left( {4x + 1} \right)}^3}}}} \cr
& {\text{by the definition of an improper integral}}{\text{}} \cr
& \int_0^\infty {\frac{{dx}}{{{{\left( {4x + 1} \right)}^3}}}} = \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{1}{{{{\left( {4x + 1} \right)}^3}}}} dx \cr
& \cr
& {\text{integrating }}\int {\frac{1}{{{{\left( {4x + 1} \right)}^3}}}} dx \cr
& {\text{set }}u = 4x + 1{\text{ then }}\frac{{du}}{{dx}} = 4,\,\,\,\,\frac{{du}}{4} = dx \cr
& \int {\frac{1}{{{{\left( {4x + 1} \right)}^3}}}} dx = \int {\frac{1}{{{u^3}}}} \left( {\frac{{du}}{4}} \right) = \frac{1}{4}\int {{u^{ - 3}}} du \cr
& = \frac{1}{4}\left( {\frac{{{u^{ - 2}}}}{{ - 2}}} \right) + C \cr
& = - \frac{1}{{8{u^2}}} + C \cr
& {\text{replace }}u = 4x + 1 \cr
& = - \frac{1}{{8{{\left( {4x + 1} \right)}^2}}} + C \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{1}{{{{\left( {4x + 1} \right)}^3}}}} dx = \mathop {\lim }\limits_{b \to \infty } \left( { - \frac{1}{{8{{\left( {4x + 1} \right)}^2}}}} \right)_0^b \cr
& = - \frac{1}{8}\mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{{{{\left( {4x + 1} \right)}^2}}}} \right)_0^b \cr
& {\text{ using ftc}} \cr
& = - \frac{1}{8}\mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{{{{\left( {4b + 1} \right)}^2}}} - \frac{1}{{{{\left( {4\left( 0 \right) + 1} \right)}^2}}}} \right) \cr
& = - \frac{1}{8}\mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{{{{\left( {4b + 1} \right)}^2}}} - 1} \right) \cr
& {\text{evaluating the limit when }}b \to \infty \cr
& = - \frac{1}{8}\left( {\frac{1}{{{{\left( \infty \right)}^2}}} - 1} \right) \cr
& = \frac{1}{8} \cr} $$
