Answer
$$\frac{1}{3}{\tan ^{ - 1}}\left( {\frac{{x + 1}}{3}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2} + 2x + 10}}} \cr
& {\text{completing the square}} \cr
& = \int {\frac{{dx}}{{{x^2} + 2x + 1 + 9}}} = \int {\frac{{dx}}{{\left( {{x^2} + 2x + 1} \right) + 9}}} \cr
& = \int {\frac{{dx}}{{{{\left( {x + 1} \right)}^2} + {{\left( 3 \right)}^2}}}} \cr
& {\text{substitute }}u = x + 1,{\text{ }}du = dx,{\text{ }}a = 3 \cr
& = \int {\frac{{du}}{{{u^2} + {a^2}}}} \cr
& {\text{a matching integral in a table of integrals at the end of the book is the }} \cr
& {\text{reduction formula 14}} \cr
& = \int {\frac{{du}}{{{u^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& u = x + 1,{\text{ }}a = 3 \cr
& = \frac{1}{3}{\tan ^{ - 1}}\left( {\frac{{x + 1}}{3}} \right) + C \cr} $$
