Answer
$$\frac{1}{3}{\tan ^{ - 1}}\left( {3x + 1} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{1 + {{\left( {3x + 1} \right)}^2}}}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = 3x + 1,{\text{ so that }}du = 3dx \cr
& {\text{then}} \cr
& \int {\frac{{dx}}{{1 + {{\left( {3x + 1} \right)}^2}}}} = \int {\frac{{du/3}}{{1 + {u^2}}}} \cr
& = \frac{1}{3}\int {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{intgrate by using the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 1 \cr
& = \frac{1}{3}{\tan ^{ - 1}}\left( {\frac{u}{1}} \right) + C \cr
& {\text{write in terms of }}x;{\text{ replace }}3x + 1{\text{ for }}u \cr
& = \frac{1}{3}{\tan ^{ - 1}}\left( {3x + 1} \right) + C \cr} $$
