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