Answer
$$\frac{1}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {\frac{{dx}}{{{{\left( {3x + 1} \right)}^2}}}} \cr
& {\text{by the definition of an improper integral}}{\text{,}} \cr
& \int_0^\infty {\frac{{dx}}{{{{\left( {3x + 1} \right)}^2}}}} = \mathop {\lim }\limits_{b \to \infty } \int_{10}^b {\frac{{dx}}{{{{\left( {3x + 1} \right)}^2}}}} dx \cr
& {\text{rewrite the integrand as}} \cr
& = \frac{1}{3}\mathop {\lim }\limits_{b \to \infty } \int_0^b {{{\left( {3x + 1} \right)}^{ - 2}}\left( 3 \right)} dx \cr
& {\text{using the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& = \frac{1}{3}\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{{\left( {3x + 1} \right)}^{ - 1}}}}{{ - 1}}} \right]_0^b \cr
& = - \frac{1}{3}\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{3x + 1}}} \right]_0^b \cr
& {\text{property of integrals}} \cr
& = \frac{1}{3}\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{3x + 1}}} \right]_b^0 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\left( {{\text{see page 388}}} \right) \cr
& = \frac{1}{3}\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{3\left( 0 \right) + 1}} - \frac{1}{{3\left( b \right) + 1}}} \right] \cr
& = \frac{1}{3}\mathop {\lim }\limits_{b \to \infty } \left( {1 - \frac{3}{b}} \right) \cr
& {\text{evaluate the limit when }}b \to \infty \cr
& = \frac{1}{3}\left( {1 - \frac{3}{\infty }} \right) \cr
& = \frac{1}{3}\left( 1 \right) \cr
& = \frac{1}{3} \cr} $$
