Answer
$${\tan ^{ - 1}}\left( {y + 3} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dy}}{{{y^2} + 6y + 10}}} \cr
& {\text{complete the square for }}{y^2} + 6y + 10 \cr
& = {y^2} + 6y + 9 + 1 \cr
& = {\left( {y + 3} \right)^2} + 1 \cr
& {\text{then}} \cr
& \int {\frac{{dy}}{{{y^2} + 6y + 10}}} = \int {\frac{{dy}}{{{{\left( {y + 3} \right)}^2} + 1}}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = y + 3,{\text{ so that }}du = dy \cr
& \int {\frac{{dy}}{{{{\left( {y + 3} \right)}^2} + 1}}} = \int {\frac{{du}}{{{u^2} + 1}}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{{u^2} + {a^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 1 \cr
& = {\tan ^{ - 1}}\left( u \right) + C \cr
& {\text{write in terms of }}y;{\text{ replace }}y - 1{\text{ for }}u \cr
& = {\tan ^{ - 1}}\left( {y + 3} \right) + C \cr} $$
